*:.'-•:.'  •'■'■--;      "...  H8SH8 


■>.■'.';:.«'■:;■:■•'■•■....■''■-■. 


ANALYTICAL  MECHANICS 

FOR   STUDENTS   OF   PHYSICS   AND 
ENGINEERING 


HAROUTUNE  M.  PADOURIAX.   M.A.,   Ph.D. 

Instructor  of  Physics  in  the  Sheffield  Scientific  School 
of  Yale  University 


™*T — ns^—r^Wf 


NEW    STORK 

D.  VAN  NOSTRAND  COMPANY 

25  Pai;k   Place 

1913 


Copyright,  1913, 

BY 

D.  VAN  NOSTRAND  COMPANY 


Stanbopc  lpreaa 

F.    H.  CILSON     COMPAN1 
BOSTON.     U.S. A 


PREFACE. 


The  following  work  is  based  upon  a  course  of  lectures  and 
recitations  which  the  author  has  given,  during  the  last  few 
years,  to  the  Junior  class  of  the  Electrical  Engineering 
Department  of  the  Sheffield  Scientific  School. 

It  has  been  the  author's  aim  to  present  the  subject  in  such 
a  manner  as  to  enable  the  student  to  acquire  a  firm  grasp  of 
the  fundamental  principles  of  Mechanics  and  to  apply  them 
to  problems- with  the  minimum  amount  of  mental  effort. 
In  other  words  economy  of  thought  is  the  goal  at  which  the 
author  has  aimed.  It  should  not  be  understood,  however, 
that  the  author  has  been  led  by  the  tendency  toward  reducing 
text-books  to  collections  of  rules,  mnemonic  forms,  and  formu- 
lae. Rules  and  drill  methods  tend  toward  the  exclusion  of 
reasoning  rather  than  toward  efficiency  in  thinking.  The 
following  features  of  the  treatment  of  the  subject  may  be 
noted : 

In  order  to  make  the  book  suitable  for  the  purposes  of 
more  than  one  class  of  students  more  special  topics  are 
discussed  than  any  one  class  will  probably  take  up.  But 
these  are  so  arranged  as  to  permit  the  omission  of  one  or 
more  without  breaking  the  logical  continuity  of  the  subject. 

In  deciding  on  the  order  of  the  topics  discussed  two  more 
or  less  conflicting  factors  have  been  kept  before  the  eye, 
i.e.,  to  make  the  treatment  logical,  yet  to  introduce  as  few 
new  concepts  at  a  time  as  possible.  It  is  to  secure  the 
second  of  these  ends,  for  instance,  that  the  historical  order 
of  the  development  of  mechanics  is  followed  by  discussing 
equilibrium   before   motion.     This   arrangement    not    only 


27105:} 


IV  PREFACE 

grades  the  path  of  the  student  by  leading  him  from  the 
easier  to  the  more  difficult  dynamical  ideas,  but  it  also 
gives  him  time  to  acquire  proficiency  in  the  use  of  his  mathe- 
matical tools. 

As  a  result  of  the  severe  criticisms  of  Newton's  laws  of 
motion  by  such  men  as  Heinrich  Hertz,  Ernst  Mach,  and 
Kail  Pearson,  authors  of  recent  text-books  on  Mechanics 
have  shown  a  tendency  to  give  either  a  new  set  of  laws  or 
Done  at  all.  There  is  no  doubt  that  a  subject  like  Mechanics 
should  start,  as  in  the  case  of  Thermodynamics,  with  a  few 
simple  laws  and  the  entire  structure  of  the  science  should  be 
based  upon  them.  In  the  present  work  the  following  law 
is  made  the  basis  of  the  entire  subject: 

To  every  action  there  is  an  equal  and  opposite  reaction,  or, 
the  sum  of  all  the  actions  to  which  a  body  or  a  part  of  a  body 
.is  subject  at  any  instant  vanishes. 

Four  concepts  are  associated  with  the  term  act  ion,  namely, 
the  concepts  of  force,  torque,  linear  kinetic  reaction,  and 
angular  kinetic  reaction.  These  are  introduced  one  at  a 
time  and  in  connection  with  the  application  of  the  law  to 
a  certain  class  of  problems.  Force  is  introduced  with  the 
equilibrium  of  a  particle  (pp.  15, 10),  torque  with  the  equilib- 
rium of  a  rigid  body  (pp.  35,  39,  40),  linear  kinetic  reaction 
with  the  motion  of  a  particle  (pp.  100-106),  angular  kinetic 
reaction  with  the  motion  of  a  rigid  body  (pp.  218-221). 
Thus  by  introducing  the  concepts  of  linear  and  angular 
reactions  and  by  extending  the  meaning  of  the  term  action 
to  include  these  reactions,  the  fundamental  principle  of 
Mechanics  is  put  in  the  form  of  a  single  law,  which  is  equiva- 
lent to  Newton's  laws  of  motion  and  which  lias  the  advan- 
tages <>f  the  point  of  view  involved  in  D'Alembert's  principle. 

This  law  ha-  the  directness  and  simplicity  of  Newton's 
third  law.  so  that   the  beginner  can  easily  understand  it  and 

apply  it  to  simple  problems  of  equilibrium,  and  yel  it  admits 
of  wider  interpretation  and  application  with   the  growth  of 


PREFACE  v 

the  student's  knowledge.  By  making  this  law  the  central 
idea  of  the  entire  subject  and  by  gradually  extending  its 
interpretation  the  treatment  is  made  uniform,  coherent,  and 
progressive. 

While  appeal  is  made  to  the  student's  experience  in  in- 
troducing the  principles  of  the  conservation  of  dynamical 
energy  and  of  the  conservation  of  momentum  they  are 
shown  to  be  direct  consequences  of  the  law  of  action  and 
reaction.  The  equivalence  and  the  alternative  character 
of  the  conservation  principles  and  of  the  law  of  action  and 
reaction  are  emphasized  by  working  out  a  number  of  prob- 
lems by  the  application  of  both  the  law  and  the  principles. 

The  two  types  of  motion,  i.e.,  motion  of  translation  and 
motion  of  rotation,  are  treated  not  only  in  the  same  general 
manner,  but  are  developed  along  almost  parallel  lines. 

The  simpler  types  of  motion  which  are  generally  treated 
under  Kinematics  are  given  in  the  present  work  as  problems 
in  Dynamics.  The  author  believes  that  the  practice  of 
divesting  the  physical  character  of  the  motion  from  the 
simpler  types  and  reducing  them  to  problems  in  integration 
is  unfortunate.  On  account  of  their  freedom  from  mathe- 
matical difficulties  the  simpler  types  of  motion  are  particu- 
larly well  adapted  to  illustrate  the  principles  of  dynamics. 

In  order  to  differentiate  between  vectors  and  their  magni- 
tudes the  former  are  printed  in  the  Gothic  type. 

In  conclusion  the  author  wishes  to  express  his  obligations 
to  Mr.  Leigh  Page  for  reading  the  manuscript  and  to  Dr. 
David  D.  Leib  for  reading  the  proofs  and  to  both  for  many 
valuable  suggestions. 

H.  M.  Dadourian. 

Yale  University, 
January,  1913. 


TABLE   OF   CONTENTS. 


PAOE 

Table  of  notations xi 

Introduction 1 


Chapter  I.     Addition  and  Resolution  of  Vectors. 

Scalar  and  vector  magnitudes 3 

Addition  of  two  vectors 4 

Difference  of  two  vectors 7 

Resolution  of  vectors  into  components 8 

Resultant  of  any  number  of  vectors 9 


Chapter  II.     Equilibrium  of  a  Particle. 

Particle 14 

Force 14 

The  law  of  action  and  reaction 15 

Condition  of  equilibrium  of  a  particle 16 

Sliding  friction ' 21 

Resultant  of  a  system  of  forces 25 

Chapter  III.     Equilibrium  of  a  Rigid  Body. 

Rigid  body 31 

Theorems  on  the  motion  of  rigid  bodies 31 

Linear  and  angular  action 35 

The  law  of  action  and  reaction 39 

Conditions  of  equilibrium  of  a  rigid  body 40 

Resultant  of  a  system  of  forces 47 

Friction  on  journals  and  pivots 51 

Rolling  friction 56 

Chapter  IV.     Equilibrium  of  Flexible  Cords. 

Simplification  of  problems 61 

Suspension  bridge  problem 62 

The  catenary 63 

Friction  belts 67 

vii 


viii  TABLE  "I    <  ONTENTS 


(  a Ai-ii.K  V.     Motion. 


PAGE 


An.ilv.~i--  of  [notion 73 

Fundamental  magnitudes     .  74 

Velocitj  77 

Angular  velocity    ...  86 

Aoodec  itioo  89 

Angular  acceleration  97 


(  ii  u'n.ii  VI.     Motion  <n   a  Pabticlb. 

Kinetic  reaction  100 

Tin-  law  hi'  action  and  reaction 101 

I  quation         100 

Difference  between  masfl  and  weight 109 

Mnt  Km  of  a  particle  under  a  constanl  force 113 

Motion  of  ;i  particle  under  :i  variable  force 125 


ChAPTEB  VII.      CXNTEBOl    MASS    1ND  MOMENT  OF  INKKTIA. 

<  !enter  of  man      140 

Momenl  of  inertia 152 


Chafteb  VIII.    Work. 

Work  done  by  a  force      164 

Work  iluiic  by  a  torque 169 

Hooke'a  law 172 

Virtual  work 180 


<   ii  \rri.i;    I  X.      EnBBGY. 

■  >f  work 185 

Kinetic  energy  of  translation 186 

Kinetic  energy  <>f  rotation 188 

191 

Potential  energy [93 

*  it  urn  of  energy 196 


CHAPTEB     \  I'll. I. H-     ..I       PORCI       \\H     \l\\T<i\l\\     POTBNTIAL. 

203 

I  inn  of  pnorgy 203 

irticle  in  a  field  of  force 204 

bj  a  bod)  m  :i  Geld  of  force 205 


TABLE  OF  CONTENTS  ix 

PAOI 

Now  conditions  of  equilibrium 207 

Stability  of  equilibrium 208 

Newtonian  potential 211 

Field  intensity 212 

Chapter  XI.     Cm  planar  Motion  of  a  Hiam  Body. 

Angular  kinetic  reaction 218 

The  law  of  act  ion  and  reaction 21S 

Experimental  definition  of  moment  of  inertia 220 

Torque  equation 221 

Comparison 222 

Torque  and  energy  methods 223 

Motion  about  a  fixed  axis L'_'  1 

Motion  about  instantaneous  axes 228 

Chapter  XII.     Impulse  and  Momentum. 

Impulse 238 

Momentum 239 

Conservation  of  momentum -II 

Motion  of  center  of  mass 242 

Central  collision 244 

Impact 249 

Efficiency  of  a  blow 251 

Motion  where  mass  varies 253 

Oblique  impact 258 

Chapter  XIII.     Angular  Impulse  and  Angular  Momentum. 

Angular  impulse 265 

Angular  momentum 266 

Conservation  of  angular  momentum 268 

Ballist  ic  pendulum 271 

Motion  relative  to  the  center  of  mass 272 

Reaction  of  axis 276 

( 'enter  of  percussion 279 

Ch  mtkr  XIV.    Motion  of  a  Particle  in  a  Central  Field  oi   Fo»  e. 

Cent  ral  field  of  force 283 

Equations  of  motion 283 

Motion  of  two  gravitating  part  icles 287 

Types  of  orbits 

Mass  of  a  planet  which  has  a  satellite 293 

Kepler's  laws 294 


x  TABLE  OF  CONTENTS 

Chaptbb  XV.     Periodic  Motion.  paqb 

Bimple  harmonic  motion 297 

Composition  of  simple  harmonic  motions  of  equal  period 303 

Elliptic  harmonic  mol  ion 306 

Pendulums 308 

I):uii|M-«l  harmonic  motion 320 

Vibrations  about  a  position  of  equilibrium 325 

Appendix  a.    Mathematical  Formulsb. 

Algebraic  relations  337 

Trigonometric  relations 338 

Exponential  and  hyperbolic  relations 339 

Appendix  B.     Mathematical  Tables. 

Logarithms  of  numbers 343 

Natural  trigonometric  functions 345 

Exponential  functions 347 

Index 349 


TABLE    OF    NOTATIONS. 


a 

=  radius,  length,  constant. 

A  = 

area,  point,  constant. 

b 

=  radius,  length,  constant. 

B  = 

point,  constant. 

c 

=  const.,  center  of  mass. 

C  = 

point,  constant. 

</ 

=  distance. 

D  = 

distance. 

e 

=  nap.    base,    coef.    of    restitu- 

E = 

total  energy. 

tion. 

F  = 

force,  frictional  force. 

f 

=  acceleration. 

G  = 

moment  of  force  or  torque. 

9 

=  grav.  acceleration. 

H  = 

height,  force  derived  from  po- 

h 

=  vertical  height,  constant . 

tential,  angular  impulse. 

i 

=  V=I. 

I  = 

moment  of  inertia. 

k 

=  constant. 

K  = 

radius  of  gyration,  constant. 

I 

=  length,  direction  cosine. 

L  = 

length,  linear  impulse. 

m 

=  mass,  direction  cosine. 

M  = 

mass. 

n 

=  number,  direction  cosine. 

N  = 

normal  component  of  force. 

0 

=  origin,  center. 

0  = 

origin,  point. 

V 

=  pressure,  page. 

P  = 

period,  point,  power. 

Q 

=  variable  magnitude. 

Q  = 

point. 

r 

=  radius,  radius  vector. 

R  = 

total  reaction,  resultant  force. 

s 

=  strain,  length  of  curve. 

S  = 

stress. 

t 

=  time. 

T  = 

tensile  force,  kinetic  energy. 

u 

=  velocity. 

U  = 

potential  energy. 

V 

=  velocity,  volume. 

V  = 

velocity,  potential. 

w 

=  weight. 

w  = 

weight,  work. 

X 

=  variable  magnitude. 

X  = 

.r-component  of  force. 

y 

=  variable  magnitude. 

Y  - 

//-component  of  force. 

z 

=  variable  magnitude. 

Z  = 

z-component  of  force. 

ar,  0,  y,  8  =  constant  angles. 

"  V" 

=  "  sum  of  all  the  .   .    .   s." 

6 

,  <f>,  \p  =  variable  angles. 

"  =  ' 

=  "is  identical  with." 

<fi  =  angle  of  friction. 

"  = ' 

=  "approaches." 

n  =  coef.  of  friction. 

"  =  ' 

=  "equals  approximately." 

X  =  modulus  of  elasticity. 

"<' 

=  "  is  smaller  than." 

7  =  angular  acceleration. 

"« 

'  =  "is     very     small     compared 

w  =  angular  velocity. 

with." 

«  =  a  small  quantity,  or  angle. 

">* 

=  "  is  greater  than." 

a  =  surface  density. 

"» 

'=  "is     very     large    compared 

r  =  volume  density. 

with." 

=  linear  density,  radius  of 
curvature. 


!" 


1-2-3.   .  . 


Xll 


TABLE  OF  NOTATIONS 


lb. 

=  pound,  the  unit  <>f  m 

tight 

gm. 

=  gram. 

pd 

=  mass    of    a    body    which 

kg. 

=  kilogram. 

weight  one  pound. 

mm. 

=  millimeter. 

in. 

■  inch 

cm. 

=  centimeter. 

ft. 

■  foot 

in. 

=  meter. 

II  P 

=  horse  power. 

km. 

=  kilometer. 

SUM. 

■  simple  harmonic  motion. 

Letters  in  gothio  type  denote  vector  magnitudes. 

A  ilnt  on  a  letter  indicates  thai  the  latter  i>  differentiated  with  respect  to 
time. 

A  letter  with  a  bar  above  il  denotes  an  average  magnitude. 
The  letter  "i"  when  used  as  a  subecripl  denotes  "any  one  of  .  .  .  ," 
ads  for  "  any  one  of  Fit  FJt  Ft,  etc," 


ANALYTICAL    MECHANICS 


INTRODUCTION. 

1.  Scope  and  Aim  of  Mechanics.  —  Mechanics  is  the  science 
of  motion.     It  has  a  twofold  object: 

First,  to  describe  the  motions  of  bodies  and  to  interpret 
them  by  means  of  a  few  laws  and  principles,  which  are  gen- 
eralizations derived  from  observation  and  experience. 

Second,  to  predict  the  motion  of  bodies  for  all  times  when 
the  circumstances  of  the  motion  for  any  one  instant  are 
given,  in  addition  to  the  special  laws  which  govern  the 
motion. 

The  present  tendency  in  science  is  toward  regarding  all 
physical  phenomena  as  manifestations  of  motion.  Compli- 
cated and  apparently  dissimilar  phenomena  are  being  ex- 
plained by  the  interactions  and  motions  of  electrons,  atom-. 
molecules,  cells,  and  other  particles.  The  kinetic  theory  of 
heat,  the  wave  theories  of  sound  and  light,  and  the  electroD 
theory  of  electricity  are  examples  which  illustrate  the  tend- 
ency toward  a  mechanical  interpretation  of  the  physical 
universe. 

This  tendency  not  only  emphasizes  the  fundamental  im- 
portance of  the  science  of  mechanics  to  other  physical 
sciences  and  engineering  but  it  also  broadens  the  aim  ot  t la- 
science  and  makes  the  dynamical  interpretation  of  all  physi- 
cal phenomena  its  ultimate  object.  The  aim  of  elementary 
mechanics  is,  however,  very  modest  and  its  -cope  is  limited  to 
the  discussion  of  the  simplest  cases  of  motion  and  equilibrium 
which  occur  in  nature. 

l 


2  ••••••  !  I-  AI.  MECHANICS 

2.  Divisions  of  Mechanics.  —  It  is  customary  to  divide 
Mechanics  into  Kinematics  and  Dynamics.  The  former 
treats  of  the  time  and  space  relations  of  the  motions  of 
bodies  without  regard  to  the  interactions  which  cause  them. 
In  other  w«.rd-.  Kinematics  is  the  geometry  of  motion.  In 
Dynamics,  on  the  other  hand,  motion  and  equilibrium  are 
treated  as  the  results  of  interactions  between  bodies;  conse- 
quently not  only  Htm  and  spaa  enter  into  dynamical  discus- 
sions, but  also  mass}  the  third  element  of  motion. 

Dynamics  in  it-  turn  is  divided  into  Statics  and  Kindles. 
Static-  is  tli«-  mechanics  of  bodies  in  equilibrium,  while 
Kineti.-  is  the  mechanics  of  bodies  in  motion. 

Chapters  II.  III.  and  IV  of  the  present  work  arc  devoted 
to  problems  in  Btatics,  while  the  rest  of  the  book,  with  the 
exception  of  Chapters  I.  V,  and  VII,  is  given  to  discussions 
of  problems  in  kinetics.  The  subject  matter  of  Chapters  I 
and  VII  is  essentially  of  a  mathematical  nature.  In  the 
former  the  addition  and  resolution  of  vector-  are  discussed, 
while  in  the  hitter  the  ( lalculus  is  applied  to  finding  centers 
of  mass  and  moment-  of  inertia.  Chapter  V  is  devoted 
mainly  to  kinematical  problems. 


CHAPTER   I. 


Vectors  are  rep- 
The  length  of  the 
Y 


ADDITION    AND    RESOLUTION    OF   VECTORS. 

3.  Scalar  and  Vector  Magnitudes. — -Physical  magnitudes  may 
be  divided  into  two  classes  according  to  whether  they  have 
the  property  of  orientation  or  not.  Magnitudes  which 
have  direction  are  called  vectors,  while  those  which  do  not 
have  this  property  are  called  scalars.  Displacement,  veloc- 
ity, acceleration,  force,  torque,  and  momentum  are  vector 
magnitudes.  Mass,  density,  w^ork,  energy,  and  time  are 
scalars. 

4.  Graphical  Representation  of  Vectors, 
resented  by  directed  lines  or  arrows, 
directed  line  represents  the  magnitude 
of  the  vector,  while  its  direction  coin- 
cides writh   that   of  the   vector.     For 
brevity  the  directed  lines  as  well  as 
the    physical    quantities    which    they 
represent    are    called    vectors.       The 
head  and  the  tail  of  the  directed  line 
are   called,  respectively,  the  terminus 
and  the  origin  of  the  vector.     In  Fig.  1,  for  instance,  P  is 
the  origin  and  Q  the  terminus  of  the  vector  a. 

5.  Notation.  —  Vectors  wdll  be  denoted  by  letters  printed  in 
Gothic  type,  while  their  magnitudes  will  be  represented  by 
the  same  letters  printed  in  italic  type.  Thus  in  Fig.  1  the 
vector  PQ  is  denoted  by  a,  but  if  it  is  desired  to  represent 
the  length  PQ  without  regard  to  its  orientation  a  is  used. 

6.  Equal  Vectors.  —  Two  vectors  are  said  to  be  equal  it  they 
have  the  same  length  and  the  same  direction.  It  follows, 
therefore,  that  the  value  of  a  vector  is  not  changed  when  it 

3 


FlQ.   l. 


4  ANALYTICAL  MKCHANICS 

is  moved  about  without  changing  its  direction  and  magni- 
tude. 

7.  Addition  of  Two  Vectors. —  Let  the  vectors  a  and  b,  Fig.  2, 
represent  two  displacements,  then  their  sum  is  another 
vector,  c,  which  is  equivalent  to  the  given  vectors.  In 
Older  to  find  c  let  us  apply  to  a  particle  the  operations  indi- 
cated by  a  and  b.  Each  vector  displaces  the  particle  along 
its  direct  ion  through  a  distance  equal  to  its  length.     There- 


/ 


Fio.  2.  Fro.  3. 

fore  applying  a  to  the  particle  at  P,  Fig.  3,  the  particle  is 
broughl  to  the  point  Q.  Then  applying  the  operation  indi- 
cated by  b  the  particle  is  broughl  to  the  point  R.  There- 
fore  the  result  of  the  two  operations  is  a  displacement  from 
/'  to  /.'.  Bui  this  is  equivalent  to  a  single  operation  repre- 
sented by  the  vector  c.  which  has  /'  for  its  origin  and  It  for 
it-  terminus.  Therefore  c  is  called  the  Bum,  or  the  resultant, 
of  a  and  b.     This  fact  is  denoted  by  the  following  vector 

equation, 

a  +  b  =  c.  (I) 

8.  Order  of  Addition.  —  The  order  of  addition  does  not  affect 
the  result.  If  in  I'm.  •"!  the  order  of  the  operations  indicated 
by  a  and  b  is  reversed  the  particle  moves  from  P  to  Q'  and 
th.-n  to  //.  Thus  the  path  of  the  particle  i-  changed  but  not 
t  be  resultant  displacement. 

9.  Simultaneous  Operation  of  Two  Vectors.  The  operations 
indicated  by  a  and  b  may  1><-  performed  simultaneously 
without   affecting  the  final  result.     In  order  to  illustrate 


ADDITION   AND   RESOLUTION   OF  VECTORS  5 

the  simultaneous  operation  of  two  vectors  suppose  the 
particle  to  be  a  bead  on  the  wire  .1/;,  Fig.  4.  Move  the 
wire,  keeping  it  parallel  to  itself,  until  each  of  its  particles 
is  given  a  displacement  represented  by  b.  Simultaneously 
with  the  motion  of  the  wire  move  the  bead  along  the 
wire  giving  it  a  displacement  equal  to  a.     At   the  end  of 


J 

t^ 

b/ 

*<>i 

/  ^T 

b*7 

A     '    "" *~ 

"    a    '     M  B 

Fig.  4. 

these  operations  the  bead  arrives  at  the  point  R.  If  both 
the  wire  and  the  bead  are  moved  at  constant  rate-  the 
resultant  vector  c  represents  not  only  the  resulting  dis- 
placement but  also  the  path  of  the  particle. 

10.  Rules  for  Adding  Two  Vectors.— The  results  of  the  last 
three  paragraphs  furnish  us  with  the  following  methods  for 
adding  two  vectors  graphically. 

Triangle  Method.  —  Move  one  of  the  vectors,  without  changing 
its  direction,  until  its  origin  falls  upon   the  ((rutin  us  of  the 
other  vector,  then  complete  the  triangle  by  drawing  a  vector  (he 
origin  of  which  coincides  with  that  of  the  first  rector.     T> 
vector  is  the  resultant  of  the  given  vectors. 

Parallelogram  Method.  —  Move  one  of  the  rectors  until  its 
origin  falls  on  that  of  the  other  rector,  complete  the  parallelo- 
gram, and  then  draw  a  vector  which  has  tit,  common  origin  of 
the  given  vectors  for  its  origin  and  which  forms  o  diagonal  of 
the  parallelogram.  The  ru  w  vector  is  the  resultant  of  tht 
vectors. 

11.  Analytical  Expression  for  the  Resultant  of  Two  Vectors. 
—  Let  a  and  b,  Fig.  5,  be  two  vectors  and  c  their  result- 


6 


ANALYTICAL  MECHANICS 


ant.     Then,  solving  the  triangle  formed  by  these  vectors, 
we  obtain 

c2  =  a2  +  b-  -f-  2  ab  cos  <f> 
b  sin  0 


and 


tan 


a  +  b  cos  0 


(ID 
(HI) 


where  a,  b,  and  c  are  the  magnitudes  of  a,  b,  and  c,  respec- 
tively, while  0  and  0  are  the  angles  b  and  c  make  with  a. 
Equation  (II)  determines  the  magnitude  and  equation  (III) 
the  direction  of  c. 


1  i...  5. 


(n) 


(b) 


I  i...  6. 


Special  Cases,     (a)  If  a  and  b  have  the  same  direction, 
as  in   Pig.  6a,  then  <£=().      Therefore 


md 


c*  =  a*+b*+2dbf 

tan  0  =  0, 


c  =  a  +  b, 
0  =  0. 


Thus  c  has  the  same  direction  as  a  and  b,  while  its  magni- 
tude equals  the  arithmetical  sum  of  their  magnitudes. 

b    When  a  and  b  are  oppositely  directed,  as  in  Fig.  6b, 
<t>  =  t.     Therefore 

C "  -  a2  +  b-  —  2  oft,       .*.      C  =  a  —  6, 

and  tan  0  =  0,  :.      0  =  0. 


ADDITION  AND  RESOLUTION   OF  VEK  TORS  7 

Thus  the  magnitude  of  c  equals  the  algebraic  Bum  of  the 

magnitudes  of  a  and  b,  while  its  direction  is  the  same  as 
that  of  the  larger  of  the  two.  It  is  evidenl  thai  it'  the 
magnitudes  of  a  and  b  are  equal  c  vanishes.  Therefore 
two  vectors  of  equal  magnitude  and  opposite  directions  are 
the  negatives  of  each  other.  In  other  words,  ,/•/„  „  the  din <- 
Hon  of  a  vector  is  reversed  its  sign  is  changed. 

(c)  When  a  and  b  are  at  right  angles  to  each  other,  as  in 

Fig.  6c,  <j>  =  l-     Therefore 

c2  =  a2  +  b2 

and  tan  0  =  -  • 

a 

12.  Difference  of  Two  Vectors.  —  Subtraction  is  equivalent  to 
the  addition  of  a  negative  quantity.  Therefore,  to  subtract 
b  from  a  we  add  —  b  to  a.  Thus 
we  have  the  following  rule  for 
subtracting  one  vector  from  an- 
other. 

In  order  to  subtract  one  vector 
from  another  reverse  the  one  to  be 
subtracted  and  add  it  to  the  other 
vector. 

It  is  evident  from  Fig.  7  that 
the  sum  and  the  difference  of  two  vectors  form  the  diagonals 
of  the  parallelogram  determined  by  them. 

ILLUSTRATIVE   EXAMPLES. 

A  particle  is  displaced  10  cm.  X.  30°  E.,  then  Jo  cm.  E.     Find  the 
resulting  displacement. 

Representing  the  displacements  and  their  resultant  by  the  vectors  a, 

b,  and  c,  Fig.  8,  we  obtain 


ANALYTICAL  MECHANICS 

c-  =  a-  +  61  +  2  oo  cos  <> 

=  (10  cm.)8  +  (10  cm.)-  +  2  X  10  cm.  X  10  cm.  cos  (60°) 
=  300  cm.1 


tan 


10  V3  cm. 
17.3  cm.* 

h  sin  <t> 


Y 

k          b 

/                    ~> 

f                      S 

S 

a/ 

yf 

.'  ° 

~-Jd 

,X 

/.• 

0 

X 

a  +  b  cos0 

10  cm.  sin  (60°) 
10  cm. +  10  cm.  cos  (00°) 
=  J  Vs. 

.:      6  =  30°. 

Therefore  the  resultani  displacement 

is  aboul  17.:;  cm.  along  the  direction  X.  G0°  E. 

PROBLEMS. 

1.  A  vector  which  points  East   has  ;t  length  of  16  cm.,  and  another 

vector  which  points  Southeast  is  25  cm.  long.    Find  the  direction  and 
the  magnitude  of  their  sum. 

2.  Find  the  direction  and  the  magnitude  of  the  difference  of  the 
vectors  of  the  last  problem. 

3.  The  sum  of  two  vectors  is  perpendicular  to  their  difference.     Show 
that  the  vectors  are  equal  in  magnitude. 

4.  The  sum  and  the  difference  of  two  vectors  have  equal  magnitudes. 
Show  that  the  vectors  are  at  right  angles  to  each  other. 


13.  Resolution  of  Vectors  into  Compo- 
nents.—  The  projection  of  a  vector  upon  a 
line  is  called  the  component  of  the  vector 
along  thai  line.  The  vectors  a*  and  a„  in 
Fig.  9,  for  instance,  are  the  components  of 
a  along  the  .r-axis  and  the  //-axis,  respec- 
tively. The  following  relal  ions  are  evidenl 
from  the  figure  and  do  not  need  further 
explanation. 


Fig.  9. 


•  The  symbol  "      "  wil]  be  used  to  denote  approximate  equality.    There- 
should  be  read  "  equals  approximately,"  or  "equals  about,"  or 
"equals  nearly."     Bee  table  of  notations,  p.  x. 


ADDITION  AND  RESOLUTION   OF  \  I  (TORS 


a  =  a,  -f 
ax  = 

a  =  V«/ +  ay\ 


a  cos  0, ) 
a  sin  0,  ^ 


tan  0  =  — 
a. 


(n 

(VI) 

\ll 


When  a  has  components  along  all  three  axes  of  a  rectangular 
system,  Fig.  10,  the  following  equations  express  the  vector 
in  terms  of  its  components. 


Ik;.   10. 

ar+ay-f 
a  COS  ai,| 
a  COS  a2,f 

a  cos  «3.| 


I\ 


V 


a  =  Vaz2+ay2+a  VI 

where  ai,  a2,  and  a:!  are  the  angles  a  makes  with  the  coordi- 
nate axes. 

14.  Resultant  of  Any  Number  of  Vectors.  Graphical  Methods. 
—  The  resultant  of  a  number  of  vectors  a.  b.  c.  etc.,  may  be 
obtained  by  either  of  the  following  metl 

First:  move  b,  without  changing  either  it-  direction  or  its 
magnitude,  until  its  origin  falls  on  the  terminus  of  a,  then 


10 


AX  A  LYTICAL  MECHANICS 


It.,.  11. 


move  c  until  its  origin  fulls  on  the  terminus  of  b,  and  so  on 
until  all  the  vectors  are  joined.  This  gives,  in  general,  an 
open  polygon.  Then  the  resultant 
is  obtained  by  drawing  a  vector 
which  closes  the  polygon  and  which 
has  its  origin  a1  the  origin  of  a. 
The  validity  of  this  method  will  be 
seen  from  Fig.  11,  where  r  repre- 
sent-; the  resultant  vector.  Evi- 
dently the  resultanl  vanishes  when 
the  given  vectors  form  a  closed 
polygon. 

Second:  draw  a  system  of  rectangular  coordinate  axes; 
resolve  each  vector  into  components  along  the  axes;  add  the 
components  along  each  axis  geometrically,  beginning  at  the 
origin.  This  gives  the  components  of  the  required  vector. 
Then  draw  the  rectangular  parallelepiped  determined  by 
these  components.  The  resultant  is  a  vector  which  has  the 
origin  of  the  axes  for  its  origin  and  forms  a  diagonal  of  the 
parallelopiped.*  This  method  is  based  upon  the  following 
analyi ical  method. 

15.  Analytical  Method. — Expressing  the  given  vectors  and  their 

resultanl  in  terms  of  their  rectangular  components,  we  have 

a=  ar+  a„  +  a„ 

b=  bx+b„+b„ 


(1) 


r  =  rx  +  r„  +  rz 
Substituting  from  (1)  in  the  vector  equation 

r=a+b  +  c+  •  •  •  (2) 

and  collecting  the  terms  we  obtain 

r,+  rir+r.  =  (a,+  b,+  -  •  •)+(a„+b1/+  .  .  .) 

+(a,+  b.+  -  •  •)•  (3) 

But  since  the  directions  of  the  coordinate  axes  are  indepen- 

•i  the  given  vectors  are  in  the  same  plane  the  parallelopiped  reduces 
to  a  n  ctangle. 


ADDITION  AND  RESOLUTION  OF  VECTORS  11 

dent,  the  components  of  r  along  any  one  of  the  axes  must 
equal  the  sum  of  the  corresponding  components  of  th< 
vectors.    Therefore  (3)  can  be  split  into  the  following  three 
separate  equations. 

r*=  a*  +  b,+  cx+  •  •  •  ,1 

r„=  au+by+c„  +  ■  •  •,  I 

r,  =  a,  +  b,  +  c,  +  •  •  •  .  ] 

It  was  shown  in  §  11  that  when  two  vectors  are  parallel 
the  algebraic  sum  of  their  magnitudes  equals  the  magni- 


Fig.  12. 

tude  of  their  resultant.  This  result  may  be  extended  to 
any  number  of  parallel  vectors.  Therefore  we  can  put  the 
vector  equations  of  (4)  into  the  following  algebraic  forms: 

rx  =  ax+bx+cx+  •  •  •  ,1 

ry=ay+by+cy+  •  •  • , 

rz  =  as  +  6,  +  c,  +  •  •  •  .  | 
Equations  (5)  determine  r  through  the  following  relations: 


vVs2+rv2+r,2, 


COS  ct\ 


COS  a3  = 


(7) 


rx 

—  ,     COS  c*2  = 

r  /'  /' 

where  ah  a2,  and  a3  are  the  angles  r  makes  with  the  axes. 
16.    Multiplication  and  Division  of  a  Vector  by  a  Scalar. 
When  a  vector  is  multiplied  or  divided  by  a  scalar  the  result 
is  a  vector  which  has  the  same  direction  as  the  original  vec- 
tor.   If,  in  the  equation  b=  ma.  m  be  a  scalar  then  b  has  the 
same  direction  as  a  but  its  magnitude  is  m  times  thai  of  a. 


12 


ANALYTICAL  MECHANICS 


ILLUSTRATIVE   EXAMPLE. 

A  man  walks  3  miles  X.  30°  E.,  then  one  mile  1  ■:..  then  3  miles  S.  45°  E., 
then  4  miles  S.,  then  one  mile  X.  .'30°  W.     Find  his  final  position. 

Representing  the  displacements  by  vectors  we  obtain  the  graphical 
solution  given  in  Fig.  13,  where  r  represents  the  resultant  displacement. 

Y 


Fig.  13. 


In  order  to  find  r  analytically  we  first  determine  its  components.    Thus 

rx  =  [3  cos  (60°)  +  cos  (0°)  +  2  cos  (-45°)  +  4  cos  (-90°) 
+  cos  (120°)]  miles 
=  (2  +  V2)  miles 
=  3.41  miles. 
r„  =  [3  sin  (00°)  +  sin  (0°)  +  2  sin  (-45°)  +  4  sin  (-90°) 
-r-sin  (120°)]  miles 
=  (2V3  -V2-4)  miles 

=  —1.95  miles. 


.-.     r  -  Vrx2  +  V 
=  3.93  miles. 

The  direction  of  r  is  given  by  the  following  relation. 


tan0 


r„  „  -1.95 
37    1 


Therefore  the  final  position  of  the  man  is  about  3.93  miles  8.  52  .9  E. 
from  his  starting  point. 


ADDITION  AND   RESOLUTION   OF   VE<   TORS  13 

PROBLEMS. 

1.  The  resultant  of  two  vectors  which  are  at  righl  angles  to  each  other 
is  twice  the  smaller  of  the  two.     The  magnitude  of  the  smaller  \  i 

a;  find  the  magnitude  of  the  greater  vector. 

2.  In  the  preceding  problem  find  the  resultant  vector. 

3.  Find  analytically  the  sum  of  three  equal  vectors  which  point  in  the 
following  directions  —  East,  N.  30°  \Y.,  and  S.  30°  \Y. 

4.  In  the  preceding  problem  make  use  of  the  first  graphical  method. 
6.    In  problem  3  make  use  of  the  second  graphical  method. 

6.  A  vector  which  is  15  cm.  long  points  X.  30°  E.  Find  its  compo- 
nents in  the  following  directions. 

(a)N.30°W.  (c)  W.  (e)S.60°E. 

(b)  X.  60°  E.  (d)  S.  30°  W.         (0  E. 

7.  A  vector  a  is  in  the  xy-jAanc.  If  3  is  added  to  nz  and  \  to  nv  the 
direction  of  the  vector  is  not  changed  but  its  magnitude  becomes  o,  +  av. 
Find  the  magnitude  and  direction  of  a. 

8.  Three  vectors  a,  b,  and  c  lie  in  the  .ry-plane.  Find  their  resultants 
analytically,  taking  the  magnitudes  of  their  components  from  the  follow- 
ing tables: 


ax 

Cly 

bx 

by 

cz 

Cy 

(1) 

6, 

9, 

—  5, 

2 

o, 

Id. 

(2) 

-3, 

7, 

5, 

o, 

6, 

-s. 

(3) 

o, 

-10, 

8, 

5, 

3, 

_2. 

(4) 

2 

o, 

-6, 

4, 

o, 

8. 

9.    In  the  preceding  problem  make  use  of  the  second  graphical  method. 

10.  Straight  horizontal  tunnels  in  a  mine  conned  the  points  /'  .  /' . 
P3,  and  P4,  in  the  given  order.  The  length  of  each  tunnel  and  the  angle 
it  makes  with  the  meridian  are  given  in  the  following  tables.  Find  the 
lengths  and  directions  of  the  tunnels  which  have  to  he  dug  in  order  to 
connect  Pi  with  P3  and  Px  directly. 

PiP2  =  200  feet,  and  makes  30°  with  the  meridian. 
p2p3  =  loo  feet,  and  makes  120  with  the  meridian. 
PJPt  -  400  feet,  and  makes  300°  with  the  meridian. 

11.  Work  out  the  preceding  problem  by  the  First  graphical  method. 

12.  Work  out  problem  10  by  the  second  graphical  method. 

13.  Find  the  direction- and  magnitude  of  the  force  experienced  by  an 
electrical  charge  of  five  units  placed  at  one  vert  ;uila1eral  tri- 
angle due  to  two  unlike  charges  of  if)  units  each  placed  at  the  other  •. 

The  sides  of  the  triangle  are  2  cm. 


CHAPTER   II. 

EQUILIBRIUM   OF  A  PARTICLE. 
ACTION  AND  REACTION.     I  ORCE. 

17.  Particle.  —  A  body  whose  dimensions  are  negligible  is 
called  a  particle.     In  a  problem  any  body  may  be  considered 

as  a  particle  so  long  as  it  docs  not  tend  to  rotate.  Even 
when  the  body  rotates  it  may  he  considered  as  a  particle  if 
its  rotation  does  not  enter  into  the  problem.  For  instance, 
in  discussing  the  motion  of  the  earth  in  its  orbil  the  earth 
is  considered  as  a  particle,  because  its  rotation  about  its 
axis  docs  not  enter  into  the  discussion. 

18.  Degrees  of  Freedom.  The  Dumberof  independent  ways 
in  which  a  body  can  move  is  called  the  number  of  degrees  of 
freedom  of  its  motion.  It  equals  the  number  of  coordinates 
which  are  necessary  in  order  to  specify  completely  the  posi- 
tion of  th<-  body.  A  five  particle  can  move  in  three  inde- 
pendent directions,  that  is.  along  the  three  axes  of  a  system 
of  rectangular  coordinates,  therefore  it  has  three  degrees 
of  freedom.  When  the  particle  is  constrained  to  move  in  a 
plane  its  freedom  is  reduced  to  two  degrees,  because  ii  can 
move  only  in  two  independent  directions.  When  it  is  con- 
strained to  move  in  a  straight  line  it  has  only  one  degree  of 
freedom. 

19.  Force.  "While  considering  the  motion  or  the  equilibrium 
of  a  body  our  attention  is  claimed  not  only  by  that  body 
hut  also  by  other-  which  act  upon  it.  In  order  to  insure 
concentration  of  attention  problems  in  Dynamics  arc  sim- 
plified in  the  following  manner.  All  bodies  are  eliminated, 
except  the  "tie  the  motion  of  which  is  being  discussed,  and 
their  actions  upon  the  latter  are  represented  by  certain  vec- 
tor magnitude-  known  as  forces.    As  an  illustration  consider 

14 


EQUILIBRIUM  OF  A  PARTICLE  15 

the  equilibrium  of  the  shaded  part  of  the  rope  in  Fig.  1  la. 
The  shaded  part  is  acted  upon  

by  the  adjoining-  sections  of  the      T      * » — ' *      * 

rope.  Therefore  we  consider  the 
shaded  part  alone  and  represenl 
the  actions  of  the  adjoining  parts 
by  the  forces  —  F  and  F,  as  shown 
in  Fig.  14b. 

20.  Definition  of  Force.  —  Force  is  a  vector  magnitudt  which 
represents  the  action  of  one  bod//  upon  another.  The  interac- 
tion between  two  bodies  takes  place  across  an  area,  while  the 
forces  which  represent  them  are  supposed  to  be  applied  al 
one  point.  Therefore  the  introduction  of  the  idea  of  force 
presupposes  the  simplification  of  dynamical  problems  which 
is  obtained  by  considering  bodies  as  single  particles,  or  as 
a  system  of  particles. 

21.  Internal  Force.  —  A  force  which  represent-  the  action  of 
one  part  of  a  body  upon  another  part  of  the  same  body  is 
called  an  internal  force. 

22.  External  Force.  —A  force  which  represents  the  action  of 
one  body  upon  another  body  is  called  an  external  force. 

23.  Unit  Force.  —  The  engineering  unit  of  force  among 
English  speaking  people  is  the  -pound.  The  pound  is  the 
weight,  in  London,  of  a  certain  piece  of  platinum  kepi  by 
the  British  government. 

24.  The  Law  of  Action  and  Reaction.— The  fundamental  law 
of  Mechanics  is  known  as  the  law  of  action  and  miction. 
Newton  (1692-1727),  who  was  the  first  to  formulate  it.  put 
the  law  in  the  following  form. 

"To  every  action  (Inn  is  an  equal  and  opposite  reaction, 
or  the  mutual  actions  of  two  bodies  are  equal  and  oppositely 
directed." 

Let  us  apply  this  law  to  the  interaction  between  a  book 
and  the  hand  in  which  you  hold  it.  Your  hand  presses 
upward  upon  the  book  in  order  to  keep  it   from   falling, 


16  ANALYTICAL  MECHANICS 

while  the  book  presses  downward  upon  your  hand.  The 
law  states  that  the  ad  ion  of  your  hand  equals  the  reaction 
of  the  book  and  is  in  the  opposite  direction.  The  book 
reacts  upon  your  hand  because  the  earth  attracts  it.  When 
your  hand  and  the  earth  are  the  only  bodies  which  act  upon 
the  book,  the  action  of  your  hand  equals  and  is  opposite  to 
the  action  of  the  earth.  In  other  words  the  sum  of  the 
two  actions  is  nil.  Generalizing  from  this  simple  illustra- 
tion we  can  put  the  law  into  the  following  form: 

To  every  action  there  is  an  equal  and  opposite  reaction,  or 
the  sum  of  all  the  actions  to  which  a  body  or  a  part  of  a  body 
is  subject  at  any  instant  vanishes  : 

2A  =  0.*  (A) 

25.  Condition  for  the  Equilibrium  of  a  Particle.  —  The  condi- 
tion of  equilibrium  of  a  particle  is  obtained  by  replacing  the 
term  "action '"  by  the  term  "force"  in  the  last  form  of  the 
fundamental  law  and  then  Stating  it  in  the  form  of  a  condi- 
tion. Thus  —  in  order  that  a  parliclr  be  in  equilibrium  the 
sum  of  all  the  forces  which  act  upon  it  must  vanish. 

In  other  words  if  Fi,  Fa,  F:t Fri  are  the  forces  which 

act  upon  a  particle,  then  the  vector  equation 

Fi+Fa+F,+  •  •  •  +F„  =  0  (I) 

musl  be  satisfied  in  order  thai  the  particle  be  in  equilib- 
rium. Equation  (I)  is  equivalent  to  stating  thai  when  the 
forces  are  added  graphically  they  form  a  closed  polygon. 
Bui  when  the  sum  of  a  number  of  vectors  vanishes  the  sum 
of  their  components  also  vanishes.  Therefore  we  must 
have 

x,+  x,+  •  •  •  +XB=0,| 

Y,  +  Y,+  ■  •  •  +Y.-0,  (II') 

Zi  +  Za+-  •  •  +  Z»  =  0,J 

where   X..  Y  .  and    Z    are  the  Components  of   F,.*      Since  the 

vectors  in  each  of  the  equations  of    II')  are  parallel  we  can 

table  of  notations. 


EQUILIBRIUM  OF  A    PARTICLE  17 

write  them  as  algebraic  equations.  Therefore  we  have  the 
following  equations  for  the  analytical  form  of  the  condition 
of  equilibrium  of  a  particle. 

2XsX1  +  X2+  •  •  •  +  A'B  =  0,*| 

2Z  m  Zx  +  Z2  +  •  •  •  +  Zn  =  0.  ) 

The  condition  of  equilibrium  may,  therefore,  be  stated 
in  the  following  form. 

In  order  that  a  particle  be  in  equilibrium  the  algebraic  sum 
of  the  components  of  the  forces  along  each  of  the  axes  of  a  rec- 
tangular system  of  coordinates  must  vanish. 

The  following  rules  will  be  helpful  in  working  out  prob- 
lems on  the  equilibrium  of  a  particle. 

First.  Represent  the  particle  by  a  point  and  the  action  of 
each  body  which  acts  upon  it  by  a  properly  chosen 
force-vector.  Be  sure  that  all  the  bodies  which 
act  upon  the  particle  are  thus  represented. 

Second.  Set  the  sums  of  the  components  of  the  forces 
along  properly  chosen  axes  equal  to  zero. 

Third.  If  there  are  not  enough  equations  to  determine  the 
unknown  quantities,  obtain  others  from  the  geo- 
metrical connections  of  the  problem. 

Fourth.  Solve  these  equations  for  the  required  quantities. 

Fifth.  Discuss  the  results. 

-      ILLUSTRATIVE    EXAMPLES. 

1.  A  particle  suspended  by  a  string  is  pulled  aside  by  a  horizontal 
force  until  the  Btring  makes  an  angle  a  with  the  vertical.  Find  the  tensile 
force  in  the  string  and  the  magnitude  of  the  horizontal  force  in  terms  of 
the  weight  of  the  particle. 

The  particle  is  acted  upon  by  three  bodies,  namely,  the  earth,  the 
string,  and  the  body  which  exerts  the  horizontal  force.    Therefore,  we 

*  The  relation  2X  a  A'i  +X»+  •  •  «  +  A'„  Ls  not  an  equation.  It 
merely  states  that   2X  is  identical  with   and  is  an   abbreviation  f< »r  A'i  -f 

x,  +  •  •  ■  +  x„. 


18 


ANALYTIC  !AL  MECHANICS 


represent  the  actions.,!'  these  bodies  by  three  force-vectors,  W,  T,  and  F, 
Pig.  15,  ami  then  apply  the  conditions  of  equilibrium.  Setting  equal  to 
zero  the  sums  of  tin-  components  of  the  y 
forces  along  the  x-  and  y-axes,  we  gel 

EX       /■'  -  7'  sin  p:  =  0.  (a) 

EY  m  -W+Tcosa  =  0.  (b) 

Solving  equations  (a)  and  (b)  we  have 


and 


cos  a 

=  T  sin  a 
=  W  tan  a. 


Discussion.  —  When  a  =  0,  T=  W 


and  F  =  0.     When  a 


2' 


cc  and 


F=x>.  Therefore  no  finite  horizontal 
force  can  make  the  string  perfectly  hori- 
zontal. 

2.  A  uniform  bar.  of  weigh!  It'  and 
length  a,  is  suspended  in  a  horizontal 
position  by  two  Btrihgs  of  equal  length  Il<;-  1,r>- 

I.     The  lower  ends  of  the  strings  are  fastened  to  the  ends  of  the  bar  and 
the  upper  ends  to  a  peg.     Find  the  tensile  force  in  the  strings. 

The  bar  is  acted  upon  by  three  bodies,  namely  the  earth  and  the  two 

Btrings.    We  represenl  their  actions  by  the  forces  W,  Ti,and  T2,  Fig.  16a. 

The  tensile  forces  of  the  strings  act  at  the  ends  of  the  liar.     On  the  other 

hand  the  weight  is  distributed  all  along  the  rod.     But  we  may  consider 

it  a-  acting  at  the  middle  point,  as  in  Fig.  16a,  or  we  may  replace  the  rod 

W 
by  two  particles  of  weight   —  each,  as  shown  in  Fig.  bib.    In  the  last  case 

the  rigidity  of  the  bar  which  prevents  its  ends  from  coining  together  is 

represented  by  the  forces  F  and  — F. 

(  lonsidering  each  particle  separately  and  setting  equal  to  zero  the  sums 
of  the  Components  of  tin1  forces  along  the  axes,  we  obtain 
EX  sTxCosa-F   =  0, 

w 

2 
for  the  first  particle,  and 

EX  =  -TiCosa  +  F  =  0, 

11' 


2F  ■  7\  sin  a 


0, 


IT       7',  sin  « 


=  0, 


EQUILIBRIUM  OF  A  PARTICLE 

for  the  second  particle.     It  follows  from  these  equations  that 

Tx  =  7', 


L9 


2  sin  a 

_J 

Vll*-d 


w. 


Fig.  16. 

Discussion.  —  The  tensile  force  of  the  strings  increases  indefinitely 

as  their  total  length  approaches  that  of  the  bar.     On  the  other  hand  as 

the  length  of  the  strings  becomes  very  large  compared  with  thai  of  the 

11' 
bar  the  tensile  force  approaches  —  as  a  limit. 

The  problem  can  be  solved  also  by  considering  the  forces  acting  on  the 
peg,  as  shown  in  Fig.  16b. 


PROBLEMS. 

1.  Show  that  when  a  particle  is  in  equilibrium  under  the  action  of  two 
forces  the  forces  must  lie  in  the  same  straight  line. 

2.  Show  that  when  a  particle  is  in  equilibrium  under  the  action  of 
three  forces  the  forces  lie  in  the  same  plane. 


20  ANALYTICAL  MECHANICS 

3.  Find  the  horizontal  force  which  will  keep  in  equilibrium  a  weight 
of  L50  pounds  on  a  smooth  inclined  plane  which  makes  oir  with  the 
horizon. 

4.  A  ring  of  weight  W  is  suspended  by  means  of  a  string  of  length  I 
'  '«■  ends  of  which  are  attached  to  two  .onus  on  the  same  horizontal  line.' 
I  ml  foe  tensde  force  oi  the  string  If  the  distance  between  .is  ends  is  d 
Also  discuss  the  limiting  cases  in  which  /  approaches  d  or  becomes  very 
large  compared  with  it.  y 

6  A  body  of  weighl  W  is  suspended  by  two  strings  of  lengths  h  and  Z, 
The  upper  end  oi  each  string  is  attached  to  a  fixed  point  in  the  same 
horizontal  hue.  Find  the  tensile  for,,,  m  the  strings  if  thedist^e 
between  the  two  points  is  d.  e 

7.  A  particle  is  in  equilibrium  on  a  smooth  inclined  plane  under  the 
action  o  wo  equd  forces,  the  one  acting  along  the  plane  upland 
the  other  horizontally.     Find  the  inclination  of  the  plane 

8.  Apply  the  conditions  of  equilibrium  to  find  the  magnitude  and 
^- of  the  resultant  of  a  number  of  forces  actmg  upoTfpaScle 

9.  1  wo  spheres  oi  equal  radius  and  equal  weighl  an-  in  eoiiSbrium  in 
asmooth  hemtepherical  bowl;  find  the  reactioiXw^  th?^o  X^ 
and  between  the  spheres  and  the  bowl.  ' 

10.  The  ^ls  of  a  string,  60  cm.  long,  are  fastened  to  two  points  in  the 
same  horizontal  hue  and  at  a  distance  of 

10  cm.  apart:  two  weights  are  hung  from 
points  in  the  Btring  25  cm.  and  20  cm.  from 
the  ends.  Find  the  ratio  of  the  weights  if 
the  part  of  the  string  between  them  is  hori- 
zontal. 

11-  A  single  triangular  truss  of  24  feet 
span  and  .",  feet  depth  supports  a  load  of 
3tans  at  the  apex.     Find  the  forces  acting  on  the  rafters  and  the  tie 

1        'Iplanchva  force  F,  acting  horizontally;  ,.  can  also  be  kept  in 

te^ns^^^01 P«^P-^  ^  the  plane.     Express  1 1 

Fand  w""'  f°U0Win8  amingementa  "f  '"lll(^  'i'"1  the  relation  between 


EQUILIBRIUM  OF  A  PARTICLE  21 

l     ""  l  i  I |  [  i 


l/w\l        l/w\l        I  /  w\  I 

(b)  (c)  (d) 


SLIDING  FRICTION. 

26.  Frictional  Force. — Consider  the  forces  acting  upon  a  body 
which  is  in  equilibrium  on  a  rough  inclined  plane,  Fig.  17. 
The  body  is  acted  upon  by  twoi 
forces,  namely,  its  weight,  W, 
and  the  reaction  of  the  plane, 
R.  The  reaction  of  the  plane 
is  the  result  of  two  distinct 
and  independent  forces.  One 
of  these,  N,  is  perpendicular 
to  the  plane  and  is  called  the 
normal  reaction.  The  other, 
F,  is  along  the  plane  and  is 
called  the  frictional  force.  The 

normal  reaction  is  due  to  the  rigidity  of  the  plain'.  It  re- 
sists the  tendency  of  the  body  to  go  through  the  plain'.  The 
frictional  force  isdueto  the  roughness  of  the  contact  between 
the  body  and  the  plane.  It  prevents  the  body  from  ^lidin.^ 
down  the  plane. 

27.  Angle  of  Friction. —  As  we  increase  the  angle  of  elevation 
of  the  inclined  plane  a  certain  definite  angle  will  be  reached 


22  ANALYTICAL  MECHANICS 

when  the  equilibrium  is  disturbed  and  the  body  begins  to 
slide  down  the  plane.  This  angle  is  called  the  angle  of 
friction.  This  definition  for  the  angle  of  friction  does  not 
hold  when  the  body  is  acted  upon  by  other  forces  besides 
its  weight  and  the  reaction  of  the  plane.  The  following 
definition,  however,  is  valid  under  all  circumstances:  The 
angle  of  friction  equals  the  angle  which  the  total  reaction  makes 
with  the  normal  to  the  surface  of  contact  when  the  body  is  on 
the  point  of  motion. 

28.  Coefficient  of  Friction.  —  Denoting  the  angle  of  friction 
by  4>,  we  obtain 

F  =  R  sin  <f>, 

N  =  R  cos  4>. 

Therefore  F  =  N  tan  </> 

=  l*N,  (HI) 

where  y.  =  tan  </>  and  is  called  the  coefficient  of  friction. 
The  angle  of  friction  and  consequently  the  coefficient  of 
friction  are  constants  which  depend  upon  the  surfaces  in 
contact.  The  last  four  equations  hold  true  only  when  the  body 
is  on  the  point  of  motion. 

29.  Static  and  Kinetic  Friction. — The  friction  winch  comes 
into  play  is  called  static  friction  if  the  body  is  at  rest  and 
kinetic  friction  if  it  is  in  motion. 

30.  Laws  of  Friction.  —  The  following  statements,  which  are 
generalizations  derived  from  experimental  results,  bring  out 
the  important  properties  of  friction.  They  hold  true  within 
certain  limits  and  are  only  approximately  true  even  within 
these  limits. 

1.  Frictional  forces  come  into  play  only  when  a  body  is 
urged  to  move. 

2.  Frictional  forces  always  act  in  a  direction  opposite  to 
that  in  which  the  body  is  urged  to  move. 

3.  Frictional  force  is  proportional  to  the  normal  reaction, 


EQUILIBRIUM  OF  A  PARTICLE 


23 


4.  Frictional  force  is  independent  of  the  area  of  contact. 

5.  The  static  frictional  force  which  comes  into  play  is  not 

greater  than  that  which  is  necessary  t<>  keep  the  body  in 
equilibrium. 

G.   Kinetic  friction  is  smaller  than  static  friction. 

Lawrs  1  to  4 hold  true  for  both  static  and  kinetic  friction. 
The  coefficient  of  friction  between  two  bodies  depends  upon 
the  condition  of  surfaces  in  contact.  Therefore  the  value  of 
n  is  not  a  perfectly  definite  constant  for  a  given  pair  of  nib- 
stances  in  contact. 

The  values  given  in  the  following  table  are  averages  of 
values  obtained  by  several  experimenters. 


Condition  of  surfaces  in 
contact. 

Coefficient  of  friction. 

Static. 

Kinetic 

Wood  on  wood 

Wood  on  wood 

Wood  on  wood 

Heavy  rope  on  wood 

Heavy  rope  on  wood 

Cast  iron  on  cast  iron 

Cast  iron  on  cast  iron 

Casl  iron  on  oak 

Leather  on  cast  iron  

Dry 
Wet 

Polished  and  greased 
Dry 
Wei 
Dry 
( Ireased 
Wet 

.50 
.68 
.35 
.60 
.80 
.24 
.15 
.65 
.30 

36 
25 
12 

.40 
.35 
.18 
.13 

ILLUSTRATIVE   EXAMPLES, 

1.   A  body  which  is  on  a  rough  horizontal  floor  can  be  brought  to  the 
point  of  motion  by  a  force  which  makes  an  angle  a  with  the  floor.     Find 
the  reaction  of  the  floor  and  the  coefficient  <>f  friction. 
The  body  is  acted  upon  by  three  forces,  Fig.  18, 
P,  the  given  force, 
W,  the  weight  of  the  body, 
R,  the  reaction  of  the  floor. 
Replacing  R  by  its  components  F  and  N.  and  applying  the  conditions 
of  equilibrium,  we  obtain 

1\Y  ■  P  cos  a-  F  =  0, 
2F  -  /'mii  a  +  N  -  W  =  0. 
Therefore  F  =  P  cos  a, 

2V  =  W  -  P  sin  a, 


24 


ANALYTICAL  MECHANICS 


and 


R  =  VF2  +  N2 

=  VP2  +  W2  -  2  PW  sin  a. 


But  since  the  body  is  on  the  point  of  motion  the  relation  F  =  /iN  holds. 
Therefore 

=  F_  _      P  cos  a 
M      N      W-P  sin  a' 

Discussion.  —  (a)    When    a  =  0,     R 
(b)    When  a  - 1,  P  =  P  -  IF  =  0,  therefore  P  =  IF,  and  /i  is  indeter- 
minate,    (c)    When  P  =  0,  m  =  0,  and  R  =  W. 
VI 


VP2  +  W2     and    n  =  ~- 
W 


Fia.  19. 

2.    A  body  which  rests  upon  a  rough  inclined  plane  is  brought  to  the 
point  of  motion  up  the  inclined  plane  by  a  horizontal  force.     Find  ^  and  R. 
The  body  is  acted  upon  by  three  forces,  Fig.  19, 

P,  the  horizontal  imrf, 

W,  the  weight, 

R,  the  reaction  of  the  plane. 

Replacing  R  by  its  components  F  and  N,  and  taking  the  axes  along  and 
ai  right  angles  to  the  plane,  we  obtain 

I'.V      P  cos  cx  -  F  -  Wean  a  =  0, 

^F        -/'.-in  «  +  X  -  W  cos  a  =  0. 


Therefore 


/'         /'  COS  Q   -    W  B0Z1  <r, 

A       P  sin  a   I   W  cos  a, 
r  -  VF*  +  N* 
'    =  y/F1  +  IP, 


and 


EQUILIBRIUM  OF  A  PARTICLE  25 

_  F_  _  P  cos  a  —  W  sin  a 
M  ~  N  ~  P  sin  a  +  W  cos  a  ' 


p 

Discussion.  —  (a)    When    a  =  0,    m  =  --     a,1(1    R  =  W Vft*  +  1. 

(b)  When  P  =  0,  m  =  —tan  a;  therefore  a  =  —  0,  that  is,  the  inclined 
plane  must  be  tipped  in  the  opposite  direction  and  mu-i  be  given  an 
angle  of  elevation  equal  to  the  angle  of  friction  in  order  that  motion  may 
take  place  towards  the  positive  direction  of  the  x-axis. 

PROBLEMS. 

1.  A  body  which  weighs  100  pounds  is  barely  started  to  move  on  a 
rough  horizontal  plane  by  a  force  of  150  pounds  acting  in  a  direction 
making  30°  with  the  horizon.     Find  R  and  /jl. 

2.  A  body  placed  on  a  rough  inclined  plane  barely  starts  to  move 
when  acted  upon  by  a  force  equal  to  the  weight  of  the  body.  Find 
the  coefficient  of  friction,  (a)  when  the  force  is  normal  to  the  plane; 
(b)   when  it  is  parallel  to  the  plane. 

3.  A  horizontal  force  equal  to  the  weight  of  the  body  has  to  be  applied 
in  order  to  just  start  a  body  into  motion  on  a  horizontal  floor.  Find  the 
coefficient  of  friction. 

4.  A  weight  IF  rests  on  a  rough  inclined  plane,  which  makes  an  angle 
a  with  the  horizon.  Find  the  smallest  force  which  will  move  the  weight 
if  the  coefficient  of  friction  is  ft. 

6.  How  would  you  determine  experimentally  the  coefficienl  of  friction 
between  two  bodies? 

6.  A  weight  of  75  pounds  rests  on  a  rough  horizontal  floor.  Find  the 
magnitude  of  the  least  horizontal  force  which  will  move  the  body  if  the 
coefficient  of  friction  is  0.4;  also  find  the  reaction  of  the  plane. 

7.  A  particle  of  weight  W  is  in  equilibrium  on  an  inclined  plane  under 
the  action  of  a  force  F,  which  makes  the  magnitude  of  the  normal  pres- 
sure equal  IF.  The  coefficient  of  friction  is  fi  and  the  angle  of  elevation  of 
the  inclined  plane  is  a.     Find  the  magnitude  and  direction  of  the  force 

8.  An  insect  starts  from  the  highesl  point  of  a  sphere  and  crawls 
down.  Where  will  it  begin  to  slide  if  the  coefficienl  of  friction  between 
the  insect  and  the  sphere  is  Y- 

9.  The  greatest  force,  which  can  keep  a  particle  at  re>t.  acting  along 
an  inclined  plane,  equals  twice  the  least  force.  Find  the  coefficienl  of 
friction.    The  angle  of  elevation  of  the  plane  i 

31.  Resultant  of  a  System  of  Forces.  -The  resultant  of  a 
number  of  forces  which  act  upon  a  particle  is  a  force  which 


26  ANALYTICAL  MECHANICS 

is  equivalent  to  the  given  forces.  There  are  two  criteria  by 
which  this  equivalence  may  be  tested.  First :  The  resultant 
force  will  give  the  particle  the  same  motion,  when  applied 
to  it,  as  that  imparted  by  the  given  system  of  forces.  We 
cannot  use  this  test  just  now  because  we  have  not  yet 
studied  motion.  Second:  When  the  resultant  force  is  re- 
versed and  applied  to  the  particle  simultaneously  with  the 
given  forces  the  particle  remains  in  equilibrium. 

According  to  the  second  criterion,  therefore,  the  resultant, 


R,  of  the  forces  Fi,  F2,  .  .  .  ,  Fn 

must  satisfy  the  equation 

-  R+(Fj+F2  +  • 
or                          R  =  Fi+F>+  •  • 

■++Fj  =  0,j               (ir) 

Splitting  the  last  equation  into 
we  obtain 

Z  =Z!+Z2  +  • 

three  algebraic  equations, 

•  •  +  X«  1 

•  •  +  rB,                   (iv) 

•  •  +  zn,\ 

where  X<,  Y„  and  Z,  are  the  components  of  Fv. 
The  magnitude  <>f  R  is  given  by  the  relation 

R  =  Vx2+Y2+Z2,  (V) 

while  the  direction  is  obtained  from  the  following  expressions 
for  its  direction  cosines. 

\  Y  Z 

COS  ai  =  —i      COS  a2  =  ~  >      COS  a3  =  —  •  (VI') 

li  li  li 

Special   Case.  —  WTien  the  forces   he   in  the  .'//-plane  the 
z-componenl  of  each  force  equals  zero.    Therefore  we  have 

B  =  VAr2+F2,  (V) 

and  tan  6     L  (VI) 

where  S  \s  the  angle  R  makes  with  the  .r-axis. 


EQUILIBRIUM   OF  A  PARTICLE  27 

PROBLEMS. 

1.  Three  men  pull  on  a  ring.  The  Brat  man  pulls  with  a  force  of  50 
pounds  toward  the  X.  30°  W.  The  second  man  pulls  toward  the  8.  15° 
E.  with  a  force  of  75  pounds,  and  the  third  man  pulls  with  a  force  of  LOO 
pounds  toward  the  west.  Determine  the  magnitude  and  direction  of 
the  resultant  force. 

2.  Show  that  the  resultant  of  two  forces  acting  upon  a  particle  Lies 
in  the  plane  of  the  given  forces. 

3.  Show  that  the  line  of  action  of  the  resultanl  of  two  forces  Lies 
within  the  angle  made  by  the  forces. 

4.  Find  the  direction  and  magnitude  of  the  resultanl  of  three  equal 
forces  which  act  along  the  axes  of  a  rectangular  system  of  coordinates. 

GENERAL   PROBLEMS. 

1.  A  particle  is  in  equilibrium  under  the  action  of  the  forces  P.  Q,  and 
R.     Prove  that 

P        _        Q  R 

sin  (Q,  R)      sin  (P,  R)      sin  (P,  Q) ' 

where  (Q,  R),  etc.,  denote  the  angles  between  Q  and  R,  etc. 

2.  Two  particles  of  weights  \\\  and  \VZ  rest  upon  a  smooth  sphere  of 
radius  a.  The  particles  are  attached  to  the  ends  of  a  string  of  Length  I, 
which  passes  over  a  smooth  peg  vertically  above  the  center  of  the  sphere. 
If  /(  is  the  distance  between  the  peg  and  the  center  of  the  sphere,  find  1 1) 
the  position  of  equilibrium  of  the  particles,  (2)  the  tensile  force  in  the 
string,  and  (3)  the  reaction  of  the  sphere. 

3.  The  lengths  of  the  mast  and  the  boom  of  a  derrick  are  ,;  and  h 

respectively.     Supposing  the  hinges  at  the  lower  end  of  the  1 m  and  the 

pulley  at  the  upper  end  to  be  smooth,  find  the  angle  the  boom  makes 
with  the  vertical  when  a  weight  W  is  suspended  in  equilibrium. 

4.  Find  the  tensile  force  in  the  chain  ami  the  compression  in  the  boom 
of  the  preceding  problem. 

6.    Two  rings  of   weights    ll'i  and   U',  are  held   on  a   smooth   circular 
wire  in  a  vertical  plane  by  means  of  a  string  subtending  an  angli 
the  center.      Show  that  the  inclination  of  the  string  to  the  horizon  is 
given  by 

tan0=(l.    j    1(jtana. 

6.  A  bridge,  Fig.  (a),  of  60-fool  span  and  10-foot  width  has  two  queen- 
post  trusses  9  feet  deep.    Each  truss  is  divided  into  three  equal  parts  by  two 


28 


ANALYTICAL   MIX'HAXK  S 


posts.     What  arc  the  stresses  in  the  different  parts  of  the  trusses  when, 
there  is  a  load  of  150  pounds  per  square  foot  of  floor  space? 


-->* — 20 — >K 20— -»i 


Fig.  (b). 


7.  Find  the  force  in  one  of  the  members  of  the  truss  of  figure  (b). 

8.  A  weight  rests  upon  a  smooth  inclined  plane,  supported  by  two  equal 
strings  the  upper  ends  of  which  are  fastened  to  two  points  of  the  plane  in 
the  same  horizontal  line.  Find  the  tensile  force  in  the  strings  and  the 
reaction  of  the  plane. 

9.  In  the  preceding  problem  suppose  the  plan^  to  be  rough. 

10.  A  particle  is  suspended  by  a  string  which  passes  through  a  smooth 
ring  fastened  to  the  highest  point  of  a  circular  wire  in  a  vertical  plane. 
The  other  end  of  the  .string  is  attached  to  a  smooth  bead  which  is  movable 
on  the  wire.  Find  the  position  of  equilibrium  supposing  the  bead  and 
the  suspended  body  to  have  equal  weights. 

11.  A  particle  is  in  equilibrium  on  a  rough  inclined  plane  under  the 
action  of  a  force  which  acts  along  the  plane.  If  the  least  magnitude  of  the 
force  when  the  inclination  of  the  plane  is  a  equals  the  greatest  magnitude 


when  it  is  «2,  show  that  0 


J,  where  4>  is  the  angle  of  friction. 


12.  Two  weights  IF,  and  W\  rest  upon  a  rough  inclined  plane,  con- 
nected by  a  siring  which  passes  through  a  smooth  pulley  in  the  plane. 
Find  the  greatest  inclination  the  plane  can  be  given  without  disturbing 
the  equilibrium. 

13.  Two  equal  weights,  which  are  connected  by  a  string,  rest  upon  a. 
rough  inclined  plane.  If  the  direction  of  the  string  is  along  the  steepest 
slope  of  the  plane  and  if  the  coefficients  of  friction  are  fi]  and  /j-2,  find  the 

greatest  inclination  the  plane  can  be  given  without  disturbing  the  equi- 
librium. 

14.  In  the  preceding  problem  find  the  tensile  force  in  the  string. 

16.  One  end  of  a  uniform  rod  rots  upon  a  rough  peg,  while  the  other 
end  is  connected,  by  means  of  a  string,  to  a  point  in  the  horizontal  plane 
which  contain-  the  peg.     When  the  roil  is  just  on  the  point  of  motion  it 


EQUILIBRIUM  OF  A    PARTICLE  29 

is  perpendicular  to  the  string.     Show  that  2/  =  n<i,  where  /  is  the 
of  the  string,  a  that  of  the  roil,  and  /i  the  coefficient  of  friction. 

16.  A  particle  resting  upon  an  inclined  plane  i-  at  the  point  of  motion 
under  the  action  of  the  force  F,  which  acts  downward  along  the  plane.  If 
the  angle  of  elevation  of  the  plane  is  changed  from  at  to  ,,_.  and  the 
direction  of  the  force  reversed  the  particle  will  barely  start  to  move  up  the 
plane.     Express  /j.  in  terms  of  a,  and  a2. 

17.  A  string,  which  passes  over  the  vertex  of  a  rough  double  inclined 
plane,  supports  two  weights.  Show  that  the  plane  must  he  tilted  through 
an  angle  equal  to  twice  the  angle  of  friction,  in  order  to  bring  it  from  the 
position  at  which  the  particles  will  begin  to  move  in  one  direction  t,.  the 
position  at  which  they  will  begin  to  move  in  the  opposite  direction. 

18.  Three  equal  spheres  are  placed  on  a  smooth  horizontal  plane  ami 
are  kept  together  by  astring,  which  surrounds  them  in  the  plane  of  their 
centers.     If  a  fourth  equal  sphere  is  placed  on  top  of  these,  prove  that  the 

tensile  force  in  the  string  is — —  ,  where  W  is  the  weigh!  of  each  sphere. 
3v6 

19.  Three  equal  hemispheres  rest  with  their  bases  upon  a  rough  hori- 
zontal plane  and  are  in  contact  with  one  another.  What  is  I  he  least  value 
of  n  which  will  enable  them  to  support  a  smooth  sphere  of  the  same  radius 
and  material'? 

20.  If  the  center  of  gravity  of  a  rod  is  at  a  distance  a  from  one  end  ami 
b  from  the  other,  find  the  least  value  of  n  which  will  allow  it  to  rest  in 
all  positions  upon  a  rough  horizontal  ground  and  against  a  rough  vertical 
wall. 

21.  A  string,  which  is  slung  over  two  smooth  pegs  at  the  same  level, 
supports  two  bodies  of  equal  weight  W  at  the  ends,  ami  a  weight  W  at 
the  middle  by  means  of  a  smooth  ring  through  which  it  passes.  Find 
the  position  of  equilibrium  of  the  middle  weight. 


CHAPTER   III. 


EQUILIBRIUM    OF    RIGID    BODIES. 
TRANSLATION   AND  ROTATION. 

32.  Rigid  Body. —  There  are  problems  in  which  bodies 
cannot  be  treated  as  single  particles.  In  such  cases  they  are 
considered  to  be  made  up  of  a  great  number  of  discrete  par- 
ticles.  A  body  is  said  to  be  rigid  if  the  distances  between 
its  particles  remain  unchanged  whatever  the  forces  to  which 
it  may  be  subjected.  There  are  no  bodies  which  are  strictly 
rigid.  All  bodies  are  deformed  more  or  less  under  the  action 
of  forces.  But  in  mo-t  problems  discussed  in  this  book  ordi- 
nary solids  may  be  treated  as  rigid  bodies. 

33.  Motion  of  a  Rigid  Body.  —  A  rigid  body  may  have  t  wo 
distinct  types  of  motion.  When  the  body  moves  so  that  its 
particles  describe  straight  paths  it 
is  Baid  t"  have  a  motion  of  trans- 
latum.  Evidently  the  paths  of  the 
particles  are  parallel.  Fig.  20.  If 
the  particles  of  the  body  describe 
circular  path-  ii  is  said  to  have  a 
motion    of  rotation.      'Flic   plane-   of 

th<-  circh-  arc  parallel,  while  their 
center-  lie  ..ii  a  straight  line  per- 
pendicular to  these  planes,  which 
i-  called  the  axis  of  rotation.    The 

motion  of  a  flywheel  is  a  well-known  example  of  motion  of 
rotation.  Suppose  A,  Fig.  21,  to  be  a  rigid  body  which  is 
brought  from  the  position  .1  to  the  position  A'  by  a  motion 
of  rotation  about  an  axis  through  the  point  0  perpendicular 

to   the   plane  of   the   paper,    then    the   paths  of   its   particles 

30 


.;.  20. 


EQUILIBRIUM  OF  RIGID  BODIES 


:;i 


FlQ.    21. 


are  arcs  of  circles  whose  planes 
are  parallel  to  the  plane  of  the 
paper  and  whose  centers  lie 
on  the  axis  of  rotation. 

34.  Uniplanar  Motion. — 
When  a  rigid  body  moves  so 
that  each  of  its  particles  re- 
mains at  a  constant  distance 
from  a  fixed  plane  the  motion 
is  said  to  be  uniplanar.  The 
fixed  plane  is  called  the  guide  plane. 

35.  Theorem  I.  —  Uniplanar  motion  of  a  rigid  body  consists 
of  a  succession  of  infinitesimal  rotational  displacements. 

Suppose  the  rigid  body  A,  Fig.  22,  to  describe  a  uniplanar 
motion  parallel  to  the  plane  of  the  paper  and  let  A  and  A' 
be  any  two  positions  occupied  by  the  body.  Then  it  may 
be  brought  from  A  to  A'  by  a  rotational  displacement 
about  an  axis  the  position  of 
which  may  be  found  in  the  fol- 
lowing manner.  Let  P  and  Q 
be  the  positions  of  any  two 
particles  of  the  body  in  a  plane 
parallel  to  the  plane  of  the 
paper  when  the  body  is  at  the 
position  A,  and  P'  and  Q'  be  the 
positions  of  the  same  particles 
when  the  body  occupies  the  po- 
sition A'.  Then  the  desired  axis  is  perpendicular  to  the 
plane  of  the  paper  and  passes  through  the  poinl  of 
intersection  of  the  perpendicular  bisectors  of  the  lines 
PP'  and  QQ',  drawn  in  the  plain'  determined  by  these 
lines. 

Therefore  the  body  can  be  broughl  from  an\  position  .1 
to  any  other  position  A'  by  a  single  rotational  displacement. 
The  actual  motion  between  .1  and  .1'  will  be,  in  general, 


r 


32  ANALYTICAL  MECHANICS 

quite  different  from  the  simple  rotation  by  which  we  accom- 
plished the  passage  of  the  body  from  one  of  these  positions 
to  the  other.  But  the  result,  which  we  have  just  obtained, 
is  true  not  only  for  positions  which  are  separated  by  finite 
distances  but  also  for  positions  which  are  infinitely  near 
each  other.  Therefore  by  giving  the  body  infinitesimal 
rotational  displacements  about  properly  chosen  axes  it  may 
be  made  to  assume  all  the  positions  which  it  occupies  during 
its  actual  motion. 

36.  Instantaneous  Axis.  — -  As  the  body  is  made  to  occupy 
the  various  positions  of  its  actual  motion  the  axis  of  rota- 
tion moves  at  right  angles  to  itself  and  generates  a  cylin- 
der whose  elements  are  perpendicular  to  the  guide  plane. 
The  elements  of  the  cylinder  are  called  instantaneous  axes, 
because  each  acts  as  the  axis  of  rotation  at  the  instant  when 
the  body  occupies  a  certain  position.  The  curve  of  inter- 
section of  the  cylinder  and  the  guide  plane  is  called  the 
centrode. 

The  motion  of  a  cylinder  which  rolls  in  a  larger  cylinder 
is  a  simple  example  of  uniplanar  motion.  In  this  case  the 
common  element  of  contact  is  the  instantaneous  axis.  As 
the  cylinder  rolls  different  elements  of  the  fixed  cylinder 
become  the  axis  of  rotation. 

.Motion  of  translation  and  motion  of  rotation  are  special 
cases  of  uniplanar  motion.  In  motion  of  translation  the 
axis  of  rotation  is  infinitely  far  from  the  moving  body.  In 
rotation  the  cylinder  formed  by  the  instantaneous  axes 
reduces  to  a  single  line,  i.e.,  the  axis  of  rotation. 

37.  Theorem  II.  —  Rotation  about  any  axis  is  equivalent  to 
a  rotation  through  the  sann  angle  about  a  parallel  axis  and  a 
translation  in  a  direction  perpendicular  to  it. 

The  truth  of  thi-  theorem  will  be  seen  from  Fig.  23,  where 
the  rigid  body  A  is  brought  from  the  position  A  to  the  posi- 
tion .1'  by  a  single  rotation  about  :m  axis  through  the  point 
0  perpendicular  to  the  plane  of  the  paper.    This  displace- 


EQUILIBRIUM   OF  RIGID   BODIES 


33 


ment  may  be  produced  also  by  rotating  the  body  to  the 
position  A"  and  then  translating  it  to  the  position  A'. 


Fig.  23. 

PROBLEMS. 

1.  Show  that  in  theorem  II  the  order  of  the  rotation  and  of  the  trans- 
lation may  be  changed. 

2.  Show  that  the  converse  of  theorem  II  is  true. 

38.  Theorem  III.  —  The  most  general  displacement  of  a  rigid 
body  can  be  obtained  by  a  single  translation  and  a  single 
rotation. 

Let  A  and  A'  be  any  two  positions  occupied  by  the  rigid 
body  and  P  and  P'  be  the  corresponding  positions  of  any  <>n<' 


Fig.  24. 


of  its  particles.  Then  the  body  may  be  brought  from  .1  to 
A'  by  giving  it  a  motion  of  translation  which  will  bring  tin- 
particle  from  P  to  P'  and  then  rotating  the  body  about  a 
properly  chosen  axis  through   P'.     A  Bpecial  case  of  this 


34  ANALYTICAL   MECHANICS 

theorem  is  illustrated  in  Fig.  24,  where  the  direction  of  the 
translation  is  perpendicular  to  the  axis  of  rotation. 

39.  Theorem  IV.  —  The  most  general  displacerm  nt  of  a  rigid 
body  can  be  obtained  by  a  displacement  similar  to  that  of  a 
screw  in  its  nut,  that  is,  by  a  rotation  about  an  axis  and  a 
translation  along  it. 

This  theorem  states  that  the  axis  of  rotation  of  the  last 
theorem  can  be  so  chosen  that  the  translation  is  along  the 
axis  of  rotation.  In  theorem  1 1 1  lei  1'1J',  Fig.  25,  be  the  path 
of  any  point  of  the  body  )C 
described  during  the  transla-  ] 
tion  and  BB  be  the  line  about 

which   the    body    is    rotated.      >\  

Draw  CC  through  P  parallel  to       J  .^ 

BB  and  drop  the  perpendicular      |  ^^ 

P'P"  upon  CC.    The  displace-  ^^ 

ment    may    be    accomplished     ,  L^^ 
now    in  '  the    following    three       i 
stages.      First:   translate  the 
body  along  the  line  CC  until       'c 
the    point     which    was    at     /' 

arrives  at  V" .  Second:  translate  the  body  alongP"P'  until 
the  point  arrives  al  /''.  Third:  rotate  the  body  about  BB 
until  it  comes  to  the  desired  position.  But  by  theorem  II 
the  last  two  operal  ions  can  be  accomplished  by  a  single  rota- 
tion about  CC.  Therefore  the  desired  displacement  can  be  ob- 
tained l»y  a  translation  along  and  a  rotation  about  the  line  '  (  . 

Evidently  the  last  theorem  holds  for  infinitesimal  dis- 
placement- as  well  as  for  finite  displacements;  therefore 
however  complicated  the  motion  of  a  rigid  body  it   can  be 

reproduced  by  a  Buccessi f  infinitesimal  screw-displace- 

ments,  each  displacement  taking  the  body  from  one  position 
which  it  ha-  occupied  during  the  motion  to  another  position 
infinitely  near  it.  Thus  at  every  instant  of  ii >  motion  the 
rigid  body  i-  displaced  like  a  screw  in  its  nut.     In  general 


EQUILIBRIUM  OF  RIGID  BODIES 

the  pitch  and  the  direction  of  the  axis  of  the  screw-motion 
change  from  instant  to  instant.  In  the  case  of  the  motion 
of  a  screw  in  its  nut  these  do  not  change. 

Translation  and  rotation  are  special  cases  of  Bcrew- 
motion.  When  the  pitch  of  a  screw  is  made  smaller  and 
smaller  it  advances  less  and  less  during  each  revolution. 
Therefore  if  the  pitch  is  made  to  vanish  the  screw  does 
not  advance  at  all  when  it  is  rotated.  Thus  rotation  i-  a 
special  case  of  screw-motion  in  which  the  pitch  is  zero. 
On  the  other  hand  as  the  pitch  of  the  screw  is  made  greater 
and  greater  the  screw  advances  more  and  more  during  each 
revolution.  Therefore  at  the  limit  when  the  pitch  is  in- 
finitely great  the  motion  of  the  screw  becomes  a  motion  of 
translation.  Thus  translation  is  a  special  case  of  screw- 
motion  in  which  the  pitch  is  infinitely  great. 

LINEAR  AND  ANGTJLAE  ACTION.    TORQUE. 

40.  Two  Types  of  Action. — We  have  seen  that  a  rigid  body 
may  have  two  different  and  independent  types  of  motion, 
namely,  motion  of  translation  and  motion  of  rotation. 
These  motions  are  the  results  of  two  independent  and 
entirely  different  kinds  of  actions  to  which  a  rigid  body  is 
capable  of  being  subjected.  We  will  differentiate  between 
these  two  types  of  action  by  adding  the  adjectives  "linear" 
and  ''angular"  to  the  term  "action."  Tims  the  action  which 
tends  to  produce  translation  will  he  called  linear  action  and 
that  which  tends  to  produce  rotation,  angular  <uii<>u. 

41.  Torque. — The  vector  magnitude  which  represents  the 
angular  action  of  one  body  upon  another  i-  called  t<>r</ut . 

42.  Couple.  —  Although  a  single  force  i~  not  capable  of 
giving  a  rigid  body  a  motion  of  pure  rotation,  two  or  more 
external  forces  will  do  it  when  properly  applied.  The 
simplest  system  of  forces  which  is  capable  of  producing 
rotation  is  known  a-  a  couple.  It  consists  of  two  equal  and 
opposite  forces  which  are  not  in  the  same  line.  Fig.  26. 


ANALYTICAL  MECHANICS 


It  is  evident  from  Fig.  26  that  a  couple  is  capable  of 
giving  a  rigid  body  a  motion  of  rotation.  But  this  is  not 
enough  to  show  that  the  effect  produced  by  a  couple  is  the 
same  as  that  produced  by  a  torque.  We  must  show  also 
that  the  couple  is  not  capable  of  producing  a  motion  of 
translation.  Consider  the  rigid  body  A,  Fig.  27,  which  is 
acted  upon  by  a  couple.     Suppose  the  couple  did  tend  to 


Fio.  26. 


Fig.  27. 


produce  a  translation  in  a  direction  BB'.  Then  pass  through 
the  body  a  smooth  bar  of  rectangular  cross-section  in  the 
direction  of  the  supposed  motion,  so  that  the  body  is  free 
to  move  along  the  bar  but  not  free  to  rotate.  When  this 
constraint  is  imposed  upon  the  rigid  body  it  behaves  like 
a  particle  and  therefore  cannot  be  given  a  motion  by  two 
equal  and  opposite  forces.  But  since  any  motion  in  the 
direction  BB'  is  qo1  affected  by  the  presence  of  the  bar, 
the  assumption  thai  the  couple  produces  a  motion  of  trans- 
lation along  BB'  musl  be  wrong.  Hence  we  see  thai  when 
the  bar  is  taken  out  the  motion  due  to  the  couple  will  be 
one  of  pure  rotal  ion. 

43.  Measure  of  Torque.  When  a  rigid  body  is  in  equi- 
librium under  the  action  of  two  couple-  it  is  always  found 
thai  the  product  of  our  of  the  force-  of  one  couple  by  the 
distance  apart  of  the  forces  of  the  same  couple  equals  the 


EQUILIBRIUM  OF  RIGID  BODIES 


37 


corresponding  product  for  the  other  couple.  In  order,  for 
instance,  that  the  rigid  body  A,  Fig.  28,  be  in  equilibrium, 
we  must  have 

FD  =  F'D'. 
Therefore  the  product  FD  is 
the  measure  of  the  torque  of 
the  couple  formed  by  the  forces 
F  and  —  F,  the  lines  of  action 
of  which  are  separated  by  the 
distance  D.  Thus  denoting 
the  torque  of  a  couple  by  G, 
we  have 

G=FD. 

The  distance  D  is  called  the  arm  of  the  couple  and  the  plane 
of  the  forces  the  plane  of  the  couple. 

44.  Unit  Torque.  — ■  The  torque  of  a  couple  whose  forces 
are  one  pound  each  and  whose  arm  is  one  foot  is  the  unit  of 
torque.     The  symbol  for  the  unit  torque  is  the  lb.  ft. 

45.  Vector  Representation  of  Torque.  —  Torque  is  a  vector 
magnitude  and  is  represented  by  a  vector  which  is  perpen- 

F*  G- 


Fio.  28. 


(I) 


Fi<;.  20. 


dicular  to  the  plane  of  the  couple.  The  vector  points  away 
from  the  observer  when  the  couple  tends  to  rotate  the  body 
in  the  clockwise  direction  and  point-  towards  the  observer 
when  it  tends  to  rotate  the  body  in  the  counterclockwise 
direction,  Fig.  29.  In  the  first  case  the  torque  i-  considered 
to  be  negative  and  in  the  second  case  positive. 


38 


ANALYTICAL   .MECHANICS 


46.  Equal  Couples.  Two  cou- 
ples are  equal  when  the  vectors 
which  represent  their  torques  are 
equal  in  magnitude  and  have  the 
same  direction.  The  three  couples 
in  Fig.  30  are  equal  if  d  =  G2  =  G3. 

Resultant  of  two  couples  is  a 
third  couple,  whose  torque  is  the 
vector  sum  of  the  torques  of  the 
given  couples. 


PROBLEMS. 

1.  Find  the  direction  and  magnitude  of  the  resultant  torque  of  three 
couples  of  equal  magnitude  the  forces  of  which  act  along  the  edges  of  the 
bases  of  a  right  prism.     The  bases  of  the  prism  are  equilateral  triangles. 

2.  In  the  preceding  problem  let  the  forces  have  a  magnitude  of  15 
pounds  each,  the  length  of  the  prism  he  2  feet  and  the  sides  of  the  bases 

10  inches. 

3.  In  problem  1  suppose  the  prism  to  have  hexagonal  bases. 

4.  In  problem  2  BUppose  the  prism  to  be  hexagonal. 

5.  A  right  circular  cone,  of  weight  W  ami  angle  2  a,  is  placed  in  a 
circular  hole  of  radius  r,  cut  in  a  horizontal  tabic.  Assuming  the  coeffi- 
cient of  friction  between  the  cone  and  the  table  to  be  n,  find  the  least 
torque  necessary  to  rotate  the  former  about  its  axis. 

47.    Moment  of  a  Force.  —  The  most  common  method  of 

giving  a  rigid  body  a  motion  of  rotation  is  to  put  an  axle 
through  it  and  to  apply  to  it  a 

force  which  acts  in  a  plane  per- 
pendicular to  the  axle.  The 
rotation  is  produced  by  the 
couple  formed  by  the  applied 
force  ami  the  reaction  of  the 
axle.  'I'll'1  torque  due  to  the 
couple  equals  the  product  of 
tb«'  applied  force  by  the  shortest  distance  from  the  axle 
to  the  lin«'  of  action  of  the  force.     It   i-  often  more  con- 


Fio.  31. 


EQUILIBRIUM  OF  RKJID  BODIES 

venient  to  disregard  the  reaction  of  the  axle.  When  this 
is  done  the  torque  of  the  couple  is  called  the  moment  of  the 
force  applied.  Therefore  the  moment  of  a  force  about  an 
axis  equals  the  product  of  the  force  by  its  lever-arm.  The  lever- 
arm  of  a  force  is  the  shortest  distance  between  the  axis  and 
the  line  of  action  of  the  force.  In  Fig.  31  the  moment  of 
F  about  the  axis  through  the  point  0  and  perpendicular  to 
the  plane  of  the  paper  is 

G  =  Fd,  (II) 

where  d  is  the  lever-arm. 

PROBLEMS. 

1.  Prove  that  the  moment  of  a  force  about  an  axis  equals  the  moment 
of  its  component  which  lies  in  a  plane  perpendicular  to  the  axis. 

2.  Prove  that  the  sum  of  the  moments  of  the  forces  of  a  couple  about 

any  axis  perpendicular  to  the  plane  of  the  couple  is  constant  and  equals 
the  torque  of  the  couple. 

48.  Degrees  of  Freedom  of  a  Rigid  Body.  —  A  rigid  body 
may  have  a  motion  of  translation  along  each  of  the  axes  of 
a  rectangular  system  of  coordinates  and  at  the  same  time 
it  can  have  a  motion  of  rotation  about  each  of  these  axe-. 
Therefore  a  rigid  body  has  six  degrees  of  freedom,  three  of 
translation  and  three  of  rotation.  When  one  point  in  it  is 
constrained  to  move  in  a  plane  the  number  of  degn 
freedom  is  reduced  to  five.  When  the  point  i-  constrained 
to  move  in  a  straight  line  the  number  becomes  tour.  When 
the  point  is  fixed  the  body  has  only  the  three  degn 
freedom  of  rotation.  If  two  points  are  li\e<l  the  body  can 
only  rotate  about  the  line  joining  the  two  point-.  There- 
fore its  freedom  is  reduced  to  one  degree.  When  a  third 
point,  which  is  not  in  the  line  determined  by  the  other  two. 
is  fixed  the  body  cannot  move  at  all.  that  i-.  it  ha-  no 
freedom  of  motion. 

49.  The  Law  of  Action  and  Reaction.      The  law  from  which 
the  conditions  of  equilibrium  of  a  particle  were  obtained  is  a 


40  ANALYTICAL  MECHANICS 

universal  law  applicable  to  all  bodies  under  all  conditions; 
therefore  it  is  applicable  to  rigid  bodies  as  well  as  to  single 
particles.  But  since  rigid  bodies  may  be  subject  to  two 
distinct  types  of  action  the  law  may  be  stated  in  the  fol- 
lowing form. 

The  sum  of  all  the  linear  and  angular  actions  to  which 

a  body  or  a  part  of  body  is  subject  at  any  instant  vanishes: 

2(A/  +  A„)=u.  (A') 

But  since  the  two  types  of  action  are  independent  of  each 
other  the  sum  of  each  type  must  vanish  when  the  combined 
sum  vanishes.  Therefore  we  can  split  the  law  into  the  fol- 
lowing two  sections. 

To  every  linear  action  there  is  an  equal  and  opposite 
linear  reaction,  or,  the  sum  of  all  the  linear  actions  to 
which  a  body  or  a  part  of  body  is  subject  at  any  instant 
vanishes : 

ZA,  =  0.  (A,) 

To  every  angular  action  there  is  an  equal  and  opposite 
angular  reaction,  or,  the  sum  of  all  the  angular  actions 
to  which  a  body  or  a  part  of  body  is  subject  at  any  in- 
stant vanishes : 

2A„  =  0.  (A.) 

50.  Conditions  of  Equilibrium  of  a  Rigid  Body.  —  I  f  we  replace 
the  term  "linear  action  "  in  the  firsl  section  of  the  law  by  the 
\\oi<l"  force"  and  the  term  "  angular  action  "  in  the  second 
section  of  the  law  by  the  word  "torque"  we  obtain  the  two 
conditions  which  must  be  satisfied  in  order  thai  a  rigid  body 
be  in  equilibrium.  Thus,  in  order  that  a  rigid  body  be  in 
equilibrium  the  following  conditions  must  be  satisfied. 

First.  The  sum  of  all  the  forces  acting  upon  the  rigid  body 
must  vanish,  thai  is,  if  F,,  Fj,  .  .  .  F„  denote  all  the  forces 

actum  upon  the  body  then  the  vector  equation 

Fi+F,+  -  •  •  +F„  =  0  (III) 

musl  be  satisfied. 


EQUILIBRIUM  OF  RIGID  BODIES  41 

Second.  The  sum  of  all  the  torques  acting  upon  tin  rigid 
body  must  vanish,  that  is,  if  d,  G_.,  .  .  .  G„  denote  all  the 
torques  acting  upon  the  body  then  the  vector  equal  ion 

G1  +  G2+  •  •  •  +G.-0  i\ 

must  be  satisfied. 

The  following  forms  of  the  statement  of  these  two  condi- 
tions are  better  adapted  for  analysis. 

First.  The  algebraic  sum  of  the  components  of  all  the  fora  s 
along  each  of  the  axes  of  a  rectangular  system  of  coordinates 
must  vanish,  that  is, 

2I  =  Ii  +  I2+  •  •  •  +xn=0,| 

27=  Fi+r2+  .  .  .  +rB=0t  (V/) 

2Z  =  Zi  +  Z2+  .  •  •  +Zn  =  0.| 

Second.  The  algebraic  sum  of  the  components  of  all  the 
torques  about  each  of  the  axes  of  a  system  of  rectangular  coor- 
dinates must  vanish,  that  is, 

HGmmGa'  +  Qa"  +  ■  ■  ■  +  f;'"1  =  0,  | 
2GV  m  Gv'  +  Gy"  +  •  •  •  +  67  =  0,  (VI ' ) 

EG.mG.'  +  G."+  ■  •  ■  +GT  =  0.1 
51.    Coplanar  Forces. —  If  two  or  more  force-  act  in  the  same 
plane  they  are  said  to  be  coplanar.     If  a  system  of  coplanar 
forces  act  in  the  xy-ptene  then  the  conditions  of  equilibrium 
reduce  to  the  following  equation-: 

2X=Xi  +  X2+  •  •  •  +Xn=0,j 

sf  ee  ]'!+}'.,+  •  •  •  +yn=o,j 

ZG^F^h  +  F,d2  +  •  •  •  +  h\dn  =  0,  V I 

where  du  d-2,  .  .  .  ,  dn  are  the  Lever-arms  of  the  forces  F  . 
F2,  .  .  .  F„,  respectively,  about  any  axis  which  is  perpen- 
dicular to  the  plane  of  the  forces,  The  z-componenta  of 
the  forces  and  the  x-  and  y-components  of  the  moments 
vanish  identically.  Consequently  they  need  not  be  con- 
sidered. 


42  ANALYTICAL  MECHANICS 

52.  Transmissibility  of  Force.  —  A  force  which  acts  upon  a 
rigid  body  may  be  considered  to  be  applied  to  any  particle 
of  the  body  which  lies  on  the  line  of  action  of  the  force.  In 
order  to  prove  this  statement  consider  the  rigid  body  A, 
Fig.  32,  which  is  in  equilibrium  under  the  action  of  the  two 


Fig.  32. 

equal  and  opposite  forces  F  and  —  F.  Now  suppose  we 
change  the  point  of  application  of  F,  without  changing 
either  its  direction  or  its  line  of  application.  Evidently 
the  equilibrium  is  not  disturbed,  because  by  moving  F  in  its 
line  of  action  we  neither  changed  the  sum  of  the  forces  nor 
the  sum  of  their  moments  about  any  axis.  Therefore  the 
line  of  action  of  a  force  is  of  importance  and  not  its  point 
of  application. 

53.  Internal  Forces.  —  Internal  forces  do  not  affect  the  equi- 
librium of  a  rigid  body.  This  is  a  direct  consequence  of  the 
law  of  "action  and  reaction."  Since  by  definition  the  in- 
ternal forces  arc  due  to  the  interaction  between  the  particles 
of. the  system  these  forces  exist  in  equal  and  opposite  pairs, 
therefore  mutually  annul  each  other. 

ILLUSTRATIVE  EXAMPLES. 

1.  A  uniform  beam  rests  with  its  Lower  end  on  smooth  horizontal 
ground  and  its  upper  end  againsl  a  smooth  vertical  wall.  The  beam  is 
held  from  slipping  by  means  of  a  string  which  connects  the  foot  of  the 
beam  with  the  foot  of  the  wall.  Find  the  tensile  force  in  the  string  and 
the  reactions  at  the  ends  of  the  beam. 

There  are  four  forces  acting  upon  the  beam,  i.e.,  the  two  reactions,  Rt 
mid  R  .  the  tensile  force  T  and  the  weight  W.  Since  both  the  ground  and 
the  wall  are  supposed  to  I  ,c  .-mooth.  Ri  is  normal  to  the  ground,  and  R2 


EQUILIBRIUM  OF  RIGID  BODIES 


43 


to  the  wall.    Therefore  denoting  the  lengths  of  the  beam  and  the  Btring 

by  I  and  a,  respectively,  we  have 
SZ  -  Rt  -  T  =  0, 
IT  =  fij  -  W=  0, 

2<70<  =-R2l  sin  a  +  W  ,'  cos  a  =  0, 

where   -Gv  denotes  the  sum  of   the  mo- 
ments of  the  forces  about  an  axis  through 
the  point  0'  perpendicular  to  the  .r//-plane. 
Solving  the  last  three  equations  we  have 
Ri  =  W, 


and 


Discussion.  —  It  should  be  noticed  that  in  taking  the  moments  the 
axis  was  chosen  through  the  point  0'  in  order  to  eliminate  the  momenta 
of  as  many  forces  as  possible  and  thus  to  obtain  a  simple  equation. 

The  reaction  Ri  is  independent  of  the  angular  position  of  the  beam 

and  equals  the  weight  W.     On  the  other  hand  Rj  and  T  vary  with  a. 

As  a  is  diminished  from  -  toO,  Rj 


Ri  =  y  cot  a 
W         a 

W     \ 

a 

tRl 

2    VZ2-a2 

f  _  W        " 

0 

Fia.  33. 

0' 

2    VI2 -a1 

When  a 


both  R2  and  T  vanish. 


and  T  increase  indefinitely. 

2.  A  ladder  rests  on  rough  horizontal  ground  and  againsi  a  rough 
vertical  wall.  The  coefficient  of  friction  between  the  ladder  and  the 
ground  is  the  same  as  that  between  the  ladder  and  the  wall.  Find  the 
smallest  angle  the  ladder  can  make  with  the  horizon  without  slipping. 

There  are  three  forces  acting  on  the  ladder,  i.e..  its  own  weighl  W  and 
the  two  reactions  Ri  and  R2.    Replacing  Ri  and  R:  by  their  components 
and  writing  the  equations  of  equilibrium  we  obtain 
SX  a  F}  -  A',  =  0, 

sy  -  Xt  +  Fz  -  w  =  o, 


2G0'=  F2l  cos  a  +  Nil  >in  a 


cos  a  =  0. 


lere  a  is  the  required 

angle. 

We  have  further 

-ft 

F, 

A. 

44 


ANALYTICAL  MECHANICS 


Solving  these  we  get 


Fx  =  r-£-t  w, 

1  +  M" 


Fig.  34. 


Discussion.  —  The  last  expression  gives  the  value  of  a  for  a  given 
value  of  fj..     When  /x  =  1,  a  =  0,  therefore  in  this  case  the  ladder  will  be 

in  equilibrium  at  any  angle  between  0  and  -  with  the  ground.     Evidently 

this  is  true  for  any  value  of  n  greater  than  unity. 

3.  Find  the  smallest  force  which,  when  applied  at  the  center  of  a 
carriage  wheel  of  radius  a,  will  drag  it  over  an  obstacle. 

The  forces  acting  on  the  wheel  are:  its  weight  W,  the  required  force  F, 
and  the  reaction  R.  Since  the  firsl  two  meet  at  the  center  of  the  wheel, 
the  direction  of  R  must  pass  through  the  center  also.  Take  the  coordinate 
axes  along  and  at  righ1  angles  to  R,  as  shown  in  Fig.  35,  and  let  F  make 
an  angle  6  with  the  .r-axis.     Then  the  equations  of  equilibrium  become 

2  X     -F  cos  6  -  A'  +  W  cos  a  =  0, 
ZF  ■  Pain  0  -  IF. sin  a  =  0, 
!<;„■     W  -a sin  a  -  Frinfl-a  =  0. 

From  either  of  the  lasi  two  equations  we  get 

r,       sin  a  tit 


Since  ll"  and  a  are  fixed  F  can  be  changed  only  by  changing  0.    Therefore 
the  minimum  value  of  F  is  given  by  the  maximum  value  of  ain  6,  i.e., 


2' 


which  makes 


Wean 


EQUILIBRIUM   OF  \IK\1D  BODIES  i:> 

a  —  h 


From  the  figure  we  obtain  cos  a  = 


therefore 


a=  -Vh(2a-h), 


w     , 

and  F  =  —  Vh  {2  a-  h). 


Fig.  35. 

Since  cos  6  =  0  the  first  equation  of  equilibrium  gives 
R  =  W  cos  a 

a 

Discussion.  —  It  will  be  observed  that  the  first  two  of  the  equations 
of  equilibrium  are  sufficient  to  solve  the  problem. 

When  h  is  zero,  F  =  0  and  R  =  W.  On  the  other  hand  when  h  =  a, 
F  =  W  and  R  =  0. 

PROBLEMS. 

1.  Prove  that  the  true  weighl  of  a  body  is  the  geometric  mean  between 
the  apparent  weights  obtained  by  weighing  it  in  both  pans  of  a  false 
balance. 

2.  A  uniform  bar  weighing  l<>  pounds  is  supported  at  the  ei 
weight  of  25  pounds  is  suspended  fnim  a  point  20  cm.  from  one  end. 
Find  the  pressure  at  the  supports  if  the  length  of  the  bar  is  50  cm. 

3.  A  uniform  rod  which  rests  on  a  rough  horizontal  Boor  and  againsl 
a  smooth  vertical  wall  is  on  the  point  of  slipping.  Find  the  reactions  at 
the  two  ends  of  the  rod. 


46  ANALYTICAL  MECHANICS 

4.  A  body  is  suspended  from  the  middle  of  a  uniform  rod  which  passes 
over  two  fixed  supports  u  feet  apart.  In  moving  the  body  G  inches  nearer 
to  one  of  the  supports  the  pressure  on  the  support  increases  by  100 
pounds.  What  is  the  weight  of  the  body  if  o  pounds  is  the  weight  of 
the  rod? 

5.  A  uniform  rod  of  length  a  and  weight  11'  is  suspended  by  two  strings 
having  lengths  U  and  /•_..  The  lower  ends  of  the  strings  are  attached  to  the 
ends  of  the  rod,  while  the  upper  ends  arc  tied  to  a  peg.  Find  the  tensile 
force  in  the  strings. 

6.  A  safety  valve  consists  of  a  cylinder  with  a  plunger  attached  to  a 
uniform  bar  hinged  at  one  end.  The  plunger  has  a  diameter  of  J  inch 
and  is  attached  to  the  bar  at  a  distance  of  1  inch  from  the  hinge.  The 
bar  is  2  feet  long  and  weighs  1  pound.  How  far  from  the  hinge  must  a 
slide-weight  of  2  pounds  be  set  if  the  steam  is  to  blow  off  at  120  pounds 
per  Bquare  inch? 

7.  The  two  legs  of  a  stepladder  are  hinged  at  the  top  and  connected 
at  the  middle  by  a  string  of  negligible  mass.  Find  the  tensile  force  in  the 
string  and  the  pressure  on  the  hinges  when  the  ladder  stands  on  a  smooth 
plane.  The  weight  of  the  ladder  is  M',  the  length  of  its  legs  /,  and  the 
length  of  the  string  a. 

8.  A  uniform  rod  rests  on  two  smooth  inclined  planes  making  angles  of 
«i  and  aj  with  the  horizon.  Find  the  angle  which  the  rod  makes  with 
the  horizon  and  the  pressure  on  the  planes. 

9.  A  rectangular  Mock  is  placed  on  a  rough  inclined  plane  whose  in- 
clination is  gradually  increased.  If  the  block  begins  to  slide  and  to  turn 
about  its  lowest  edge  simultaneously  find  the  coefficient  of  friction. 

10.  A  uniform  rod  rests  with  one  end  against  a  rough  vertical  wall 
and  the  other  end  connected  to  a  point  in  the  wall  by  a  string  of  equal 
length.      Show  that    the  smallest  angle  which  the  string  can  make  with 

the  wall  is  tan-1  (  ■  J  • 

11.  A  uniform  rod  is  suspended  by  a  string  which  is  attached  to  the 
ends  and  is  Blung  over  a  smooth  peg.  Show  that  in  equilibrium  the  rod 
is  either  horizontal  or  vertical. 

12.  A  ladder  25  feet  lonur  and  weighing  50  pounds  rests  against  a 
vertical  wall  making  30°  with  it.  How  high  can  a  man  weighing  150 
pounds  climb  up  the  ladder  before  it  begins  to  slip'.'    The  coefficient  of 

friction  is  ()..">  at   both  ends  of  the  ladder. 

13.  A  rod  of  negligible  weight  rests  wholly  inside  a  smooth  hemispheri- 
cal bowl  of  radius  r.     A  weight    W  is  clamped  on  to  the  roil  at   a  point 

distances  from  the  ends  are  a  and  h.     Show  that   the  equilibrium 


EQUILIBRIUM   OF   RICH)    I'.oDIKS 


47 


,  where  8  is  the  angle  it 


position  of  the  rod  is  given  by  sin  6  = — - 

2Vr*-ab' 
makes  with  the  plane  of  the  brim  of  the  bowl  which  is  horizontal 

14.  Prove  that  when  a  rigid  body  is  in  equilibrium  under  the  action  of 
three  forces  their  lines  of  action  lie  in  the  same  plane  and  intersect  at  the 
same  point. 

15.  Find  the  forces  which  tend  to  compress  or  extend  the  different 
members  of  the  following  cranes. 


1,760  lbs] 


16.  Supposing  the  weights  of  the  following  figures  to  be  in  equilibrium 
find  their  relative  magnitudes.  The  circles  which  are  tangent  to  other 
circles  represent  gears. 


54.  Resultant  of  a  System  of  Forces  Acting  upon  a  Rigid  Body. 
—We  have  already  shown.tha1  the  most  general  displacement 
of  a  rigid  body  consists  of  a  translation  along,  and  a  rotation 
about,  a  certain  lino.  Therefore  such  a  displacement  can  be 
prevented  by  a  single  force  opposed  to  the  translation  and 
asingle  torque  opposed  to  the  rotation.  Thus  a  Bingle  force 
and  a  single  torque  can  be  found  which  will  keep  a  rigid  body 
in  equilibrium  against  the  action  of  any  system  of  forces. 


48  ANALYTICAL  MECHANICS 

The  resultant  of  a  system  of  forces  consists,  therefore,  of  a 
single  force  and  a  single  torque  which,  when  reversed,  will 
keep  the  rigid  body  in  equilibrium  against  the  action  of  the 
given  system  of  forces. 

55.  Resultant  of  Coplanar  Forces  Acting  upon  a  Rigid  Body.  — 
Let  Fi,  F>,  .  .  .  Fn  denote  the  given  forces  and  let  the  xy- 
plane  be  their  plane  of  action.  Then,  if  R,  X,  and  Y  denote 
the  resultant  force  and  its  components,  respectively,  we  have 

X  =  Xi  +  X2+  •  ■  •  +  xn, 


Y  =  Fi+IV+  .  .  .  +  F.  ' 

R=Vx2+Y2,  (VIII) 

Y 

and  tan  0  —  —  >  (IX) 

A 

where  the  terms  in  the  right-hand  members  of  the  first  two 
equations  are  the  components  of  the  given  forces,  and  0  is 
the  angle  R  makes  with  the  a>axis. 

On  the  other  hand  if  G0  denotes  the  resultant  torque  and 
dh  dz,  .  .  .  ,  dn  denote  the  distances  of  the  origin  from  the 
lines  of  action  of  the  forces,  then 

G.-FA+FA+  •  •  •  +Fndn.  (X) 

If  we  represent  this  torque  by  the  moment  of  the  resultant 
force  about  the  z-axis,  then 

RD  =  Fidi  +  F2<k+  ■  ■  •  +Fnd„,\ 

gives  the  distance  of  the  line  of  action  of  the  resultant  force 
from  the  origin. 

ILLUSTRATIVE  EXAMPLE. 

Find  the  resultant  of  the  six  forces  acting  along  the  sides  of  the  hexa- 
gon of  Fig.  36. 

Taking  the  sum  of  the  components  along  the  x  and  y  directions,  we 
have 


EQUILIBRIUM  OF  RIGID  BODIES 


!'.» 


X  =  2F  +  3Fco3|-2Fcos|-F-2fco8=  •  I 

=  F. 


-fVs. 


...      R  =  y/f*  +  3  p* 
=  2  F 
and  tan  8  =  —  V3. 

Therefore  the  resultant  force  has 
a  magnitude  2  F  and  makes  an  angle 
of  —60°  with  the  .r-axis. 

Taking  the  moments  about  an 
axis  through  the  center  of  the  hexa- 
gon, we  obtain 

RD  =  (2F  +  3F  +  2F  +  F  +  2F  +  F)a 
=  11  Fa, 
therefore  D  =  5.5  a, 

where  a  is  the  distance  of  the  center  from  the  lines  of  action  of  the 


Fig.  36. 


56.  Resultant  of  a  System  of  Parallel  Forces.  —  Let  R  be  the 
resultant  of  the  parallel  forces  Fi,  F2,  .  \  .  ,  F„,  which  acl 
upon  a  rigid  body.  Then,  since  the  forces  arc  parallel,  the 
resultant  force  equals  the  algebraic  sum  of  the  given  forces. 
Thus 

R=Fi  +  F2+  •  •  •  +Fn, 
and  RD  =  Fxdx  + F2d2+  ■  ■  •  +/ 

Now  take  the  z-axis  parallel  to  the  forces  and  lot  .r,  and  //, 
denote  the  distances  of  F<  from  the  //r-plano  and  ti 
plane,  respectively.     Then  the  last  equation  may  be  split 
into  two  parts,  one  of  which  gives  the  moments  about  the 
x-axis  and  the  other  about  the  //-axi-.     Thus, 

Rx  =  F1Xi  +  Fixi+  ■  ■  ■  +/ 

Ry  =  /'V/i  +  /« \!h  +  •  •  ■  +/■'„//.,  S 

where  x  and  y  are  the  coordinates  of  the  poinl  in  the  xy~ 

plane  through  which  the  resultant  force  passes.     In  other 


XII 


50 


ANALYTICAL  MECHANICS 


words,  (x,y)  is  the  point  of  application  of  the  resultant  force. 
The  resultant  force  is  evidently  parallel  to  the  given  forces. 
The  last  two  equations  may  be  written  in  the  following 
forms 

_     ZFx 


y       R 


(XIII) 


ILLUSTRATIVE   EXAMPLE. 

Find  the  resultant  of  two  parallel  forces  which  act  upon  a  rigid  body 
in  the  same  direction. 

Let    the   ?/-axis   be   parallel   to   the 
forces. 
Then  R  =  Fx  +  F2, 

a  ~      F,.r,  +  F2x2 

and  X  =     Fl  +  F2    ' 

F2      x  —  xi 

But  since  x2  —  x  and  x  —  Xi  are  the 
distances  of  F2  and  Fi  from  R,  we  have 

F\  =  d_2 

F2      ch' 
or  Fxdy  =  F'l:. 

Therefore  the  distances  of  the  resultant  from  the  given  forces  are  in- 
versely proportional  to  the  magnitudes  of  the  latter. 

PROBLEMS. 

1.  Find  the  resultant  force  and  the  resultant  torque  due  to  the  forces 
P,  2  P,  4  P  and  2  P  which  act  along  the  sides  of  a  square,  taken  in  order. 

2.  Three  forces  are  represented  in  magnitude  and  line  of  action  by 
the  Bidea  of  an  equilateral  triangle.  Find  the  resultant  force,  taking  the 
directions  of  one  of  the  forces  opposite  to  that  of  the  other  two. 

3.  The  lines  of  action  of  three  (nn-cs  form  a  right  isosceles  triangle  of 

.  0,  and  a  \  2.    The  magnitudes  of  the  forces  are  proportional  to 
the  ndes  of  the  triangle.     Find  the  resultant  force. 

4.  The  Bum  of  the  moments  of  a  system  of  coplanar  force-  about  any 

three  points,  which  arc  not  in  1  he  same  straight  line,  are  the  same.     Show 

that  the  system  is  equivalent  to  a  couple. 


EQUILIBRIUM  OF  RIGID  BODIES  51 

5.  Three  forces  are  represented  in  magnitude,  direction,  and  line  of 
action  by  the  sides  of  a  triangle  taken  in  order;  prove  thai  their  resultant 
is  a  couple  the  torque  of  which  equals,  numerically,  twice  the  area  of  the 

triangle. 

6.  Three  forces  act  along  the  sides  nf  an  equilateral  triangle;  find  the 

condition  which  will  make  their  resultant  pass  through  the  center  of  tin- 
triangle. 

FRICTION  ON  JOURNALS  AND  PIVOTS. 
57.  Friction  on  Journal  Bearings.  —  If  the  horizontal  shaft 
of  Fig.  38  fits  perfectly  in  its  bearings  the  friction  which  comes 
into  play  is  a  sliding  friction,  therefore  the  laws  of  Bliding 
friction  may  be  assumed  to  hold  good.  The  most  importanl 
of  these  laws  is:  the  frictional  force  which  comes  into  play 
is  proportional  to  the  normal  reaction,  that  is,  in  the  relation 

F  =  fiN, 

n  is  independent  of  N.  We  will  assume  therefore  thai  this 
law  holds  at  each  point  of  the  surface  of  contact  and  thus 
reduce  the  problem  under  discussion  to  one  of  sliding  fric- 
tion. There  is  an  important  difference,  however,  between 
the  problem  under  discussion  and  the  problems  on  friction 
which  we  have  already  discussed.  In  the  presenl  problem 
the  normal  reaction  is  not  the  same  at  all  the  point-  of 
the  surfaces  in  contact.  We  must  apply,  therefore,  the 
laws  of  friction  to  small  elements  of  surfaces  of  contact  over 
which  the  normal  reaction  may  be  considered  to  be  constant. 
Let  the  element  of  surface  be  a  strip,  along  the  Length  of 
the  shaft,  which  subtends  an  angle  dB  al  the  axis  of  the  shaft. 
Further  let  r/N  be  the  normal  reaction  over  this  clement  of 
surface,  and  d¥  be  the  corresponding  frictional  force;   then 

we  have 

dF  =  n  dN 

=  iip  •  I  -a  do, 

where  p  is  the  normal  reaction  per  unit  area  or  the  pressure, 
a  is  the  radius  of  the  shaft,  and  /  the  length  of  the  bearing. 


52 


ANALYTICAL  MECHANICS 


Therefore  the  total  frictional  force  and  the  total  frictional 
torque  are,  respectively, 


F=  ftal 


and 


G  =  narl 


f'pdd 
f  "p  de. 


In  order  to  carry  out  the  integral  of  the  foregoing  expressions 
we  have  to  make  some  assumption  with  regard  to  the  nature 


Fig.  38. 

of  dependence  of  p  upon  0.  But  whatever  the  relation 
between  p  and  0  it  is  obvious  that  the  sum,  over  all  the  sur- 
faces of  contact,  of  the  vertical  component  of  the  normal 
reaction  must  equal  the  load  which  rests  upon  the  bear- 
ings. If  P  denotes  this  load,  then  ;;  must  satisfy  the  condi- 
tion 

P  =  I    p  sin  0  •  dA 

Jo 

=  al  J  Tp  sin  6  <I9, 

where  .1    is  the  total  area  of  contact. 

ILLUSTRATIVE   I  XAMPLE. 
The  normal  pressure  on  the  bearings  is  given  by  the  relation  p  =  p0  sin  6; 

find  tin-  total  frictional  force  and  the  lotal  frictional  torque. 


EQUILIBRIUM  OF  RIGID  BODIES  53 

Substituting  the  given  value  of  p  in  the  expression  for  /•'  we  obtain 
F  =  nalpo  f'sin  6  dd 

Jo 

=  ->  v<dp0. 

In  order  to  determine  p0  in  terms  of  the  total  load  on  the  bearings  we 

make  p  satisfy  the  condition 


=  al  C  p  sin  6 

Jq 


60. 


Substituting  the  given  value  of  p  in  the  right-hand  member  of  the  pre- 
ceding equation  we  have 

P  =  alpo  Csm2  8  dd 

Jo 
iralpo 
~      2     ' 
2P 
Po  =  ^' 

Therefore  F  =  lif  p 

and  G  =  —  aP. 

7T 

It  will  be  observed  that  the  total  frictional  force  varies  with  the  load  and 
is  independent  of  the  radius  and  of  the  length  of  the  bearing;  in  other 
words  it  is  independent  of  the  area  of  contact. 


PROBLEMS. 

1.  Supposing  the  normal  pressure  to  be  the  same  at  every  point  of  the 
surfaces  of  contact,  derive  the  expressions  for  the  total  frictional  force  and 
the  resisting  torque  due  to  friction. 

2.  Supposing  the  vertical  componenl  of  the  total  reaction  at  every 
point  of  the  surfaces  of  contact  to  be  constant,  derive  the  expressions  for 
the  total  frictional  force  and  the  resisting  torque  due  to  friction. 

3.  Derive  expressions  for  the  total  frictional  force  and  the  resisting 
torque  upon  the  assumption  that  the  normal  pressure  is  given  by  the 
relation  p  =  p0  sin2  6. 

58.    Friction  on  Pivots.  — The    problem   of   friction    on 

pivots  also  is  a  problem  of  sliding  friction.     The  feature 


54 


ANALYTICAL   ML(  HANK'S 


which  distinguishes  the  pivot   from  the  journal  bearing  is 

this:  in  the  former  the  lever  arm  of  the  frictional  force  varies 

from  point  to  point,  while 

in  the   latter  it  is  constant 

and  equals  the  radius  of  the 

shaft. 

Let  dN  be  the  normal 
reaction  upon  dA,  an  ele- 
ment of  area  at  the  base  of 
the  flat-end  pivot  of  Fig. 
39;  then  if  dF  denotes 
the  corresponding  frictional 
force,  we  have 

dF=»dN 
=  up  dA, 

where  p  is  the  normal  pres- 
sure. Evidently  p  is  con- 
stant ;  therefore  we  can  write 


o    dA 


TTd'up. 


Fig.  39. 


The  expression  for  the  resisting  torque  due  to  the  friction 
Is  obtained  as  follows : 


G=f*r.dF 
=  I    r  •  updA 

=  fl     rup  .  rdd  •  dr 

Jo  Jo 

=  ir up  I  r2  dr 


=  f  Trasnp 
=  |  aixP, 
where  P  is  the  total  load  on  the  pivot. 


EQUILIBRIUM  OF  RIGID  BODIES 


55 


PROBLEMS. 

1.  Derive  an  expression  for  the  resisting  torque  due  to  friction  in  the 
collar-bearing  pivot  of  the  adjoining  figure. 


I' 


-2  a > 


2.  Supposing  the  normal  pressure  to  be  constant,  derive  an  expression 
for  the  resisting  torque  due  to  friction  in  the  conical  pivot  of  the  adjoining 
figure. 

3.  In  the  preceding  problem  suppose  the  vertical  component  of  the 
normal  pressure  to  be  constant. 

4.  In  problem  2  suppose  the  horizontal  component  of  the  normal 
pressure  to  be  constant. 


6.  Taking  the  normal  pressure  to  be  constant  derive  an  Depression  for 
the  resisting  torque,  due  to  friction  in  the  spherical  pivot  of  the  adjoin- 
ing figure. 


5G 


ANALYTICAL  MECHANICS 


6.  Prow  thai  the  resisting  torque  due  to  friction  is  greater  for  a  hollow 
pivot  than  for  a  solid  pivot,  provided  thai  the  load  and  the  load  per  unit 
area  arc  the  same  in  both  cases. 

7.  Show  thai  the  resisting  torque  due  to  friction  for  a  hemispherical 
pivot  is  about  2..'3.">  times  as  large  as  that  for  a  flat  end  pivot. 


ROLLING   FRICTION. 

59.  Coefficient  of  Rolling  Friction.  —  Consider  a  cylinder, 
Fig.  40,  which  is  in  equilibrium  on  a  rough  horizontal  plane 
under  the  action  of  a  force  S- 
In  addition  to  this  force  the 
cylinder  is  acted  upon  by  its 
weight  and  by  the  reaction  of 
the  plane.  Applying  the  con- 
ditions of  equilibrium  we  ob- 
tain 

27  =  -w  +  N=0, 
2GQ  =  ND-Sd=0, 

where  F  and  N  are  the  com- 
ponents of  R,  the  reaction  of 
the  plane,  while  D  and  d  are, 
respectively,  the  distances  of 
the  points  of  application  of  R  and  S  from  the  point  0,  about 
which  the  moments  are  taken.     These  equations  give  us 


Fig.  40. 


.-Hid 


R=  Vf2+N2 
=  VS-+  IP, 


It  the  cylinder  is  just  on  the  point  of  motion 

F=»N, 

s 


and  consequently 


W 


(1) 
(2) 


(3) 


EQUILIBRIUM  OF  RIGID  BODIES  57 

Combining  (2)  and  (3),  we  obtain 

D=nd.  \l\ 

The  distance  D  is  called  the  coefficient  of  rolling  friction. 
Equation  (XIV)  states,  therefore,  that  the  coefficient  of  the 
rolling  friction  equals  the  coefficient  of  the  sliding  friction 
times  the  distance  of  the  point  of  contact  from  the  line  of 
action  of  the  force  which  urges  the  body  to  roll. 

60.  Friction  Couple.  —  It  is  evident  from  the  above  equa- 
tions that  a  change  in  the  value  of  d  does  not  affect  the  values 
of  TV  and  F,  consequently  it  does  not  change  the  value  of  m- 
This  is  as  it  should  be,  since,  according  to  the  laws  of  sliding 
friction,  n  depends  only  upon  the  nature  of  the  surfaces  in 
contact.  A  change  in  d,  however,  changes  the  value  of  D; 
in  other  words,  it  changes  the  point  of  application  of  R. 
When  d  =  0,  that  is,  when  S  is  applied  at  the  point  of  con- 
tact, D  =  0,  in  which  case  the  body  is  urged  to  slide  only. 
But  when  d  is  not  zero  the  force  S  not  only  urges  the  body 
to  slide  but  also  to  roll;  therefore,  in  addition  to  the  resist- 
ing force  F,  a  resisting  torque  comes  into  play.  This  torque, 
which  is  due  to  the  couple  formed  by  N  and  W,  is  called 
the  friction  couple. 

PROBLEMS. 

1.  A  gig  is  so  constructed  that  when  the  shafts  arc  horizontal  the 
center  of  gravity  of  the  gig  is  over  the  axle  of  the  wheels.  The  gig  rests 
on  perfectly  rough  horizontal  ground.  Find  the  leasl  force  which,  act- 
ing at  the  ends  of  the  shafts,  will  just  move  the  gig. 

2.  Find  the  smallest  force  which,  acting  tangentially  at  the  rim  of  a 
flywheel,  will  rotate  it.  The  weight  and  the  radius  "f  the  flywheel,  the 
radius  of  the  shaft,  and  the  coefficienl  of  friction  between  the  shaft  and 
its  bearings  are  supposed  to  be  known. 

3.  A  flywheel  of  500  pounds  weighl  is  brought  to  the  point  of  rotation  by 

a  weight  of  10  pounds  suspended  by  means  of  a  String  wound  around  its  rim. 

Find  the  coefficienl  of  friction  between  the  axle  and  it-  bearings.    The 

diameters  of  the  wheel  and  the  axle  are  lo  feet  and  8  inches,  respectively. 


58  ANALYTICAL  MECHANICS 

4.  A  wheel  <>f  radius  a  and  weight  W  stands  on  rough  horizontal 
ground.  If  ju  is  the  coefficient  of  friction  between  the  wheel  and  the 
ground  find  the  smallest  weight  which  must  be  suspended  at  one  end 
of  the  horizontal  diameter  in  order  to  move  the  wheel. 


GENERAL   PROBLEMS. 

1.  A  table  of  negligible  weight  has  three  legs,  the  feet  forming  an 
equilateral  triangle.  Find  the  proportion  of  the  weight  carried  by  the 
legs  when  a  particle  is  placed  on  the  table. 

2.  A  rectangular  board  is  supported  in  a  vertical  position  by  two 
smooth  pegs  in  a  vertical  wall.  Show  that  if  one  of  the  diagonals  is 
parallel  to  the  line  joining  the  pegs  the  other  diagonal  is  vertical. 

3.  A  uniform  rod  rests  with  its  two  ends  on  smooth  inclined  planes 
making  angles  a  and  /3  with  the  horizon.  Where  must  a  weight  equal  to 
that  of  the  rod  be  clamped  in  order  that  the  rod  may  rest  horizontally? 

4.  A  uniform  ladder  rests  against  a  rough  vertical  wall.  Show  that 
the  least  angle  it  can  make  with  the  horizontal  floor  on  which  it  rests  is 

given  by  tan  6  =  -      MM  ,  where  ju  and  y!  are  the  coefficients  of  friction 
2  n 

for  the  floor  and  the  wall,  respectively. 

6.    A  uniform  rod  is  suspended  by  two  equal  strings  attached  to  the 

ends.     In  position  of  equilibrium  the  strings  are  parallel  and  the  bar  is 

horizontal.     Find  the  torque  which  will  turn  the  bar,  about  a  vertical 

axis,  through  an  angle  0  and  keep  it  in  equilibrium  at  that  position. 

6.  The  line  of  hinges  of  a  door  makes  an  angle  a  with  the  vertical. 
Find  the  resultant  torque  when  the  door  makes  an  angle  (3  with  its  equi- 
librium position. 

7.  The  lines  of  action  of  four  forces  form  a  quadrilateral.  If  the 
magnitude  of  the  forces  are  o,  b,  c,  d  times  the  sides  of  the  quadrilateral 
find  the  conditions  of  equilibrium. 

8.  A  force  acts  at  the  middle  point  of  each  side  of  a  plane  polygon. 
Each  force  is  proportional  to  the  length  of  the  side  it  acts  upon  and  is 
perpendicular  to  it.  Prove  that  the  polygon  will  be  in  equilibrium  if  all 
the  forces  are  directed  towards  the  inside  of  the  polygon. 

9.  A  force  acts  at  each  vertex  of  a  plane  convex  polygon  in  a  direc- 
tion parallel  to  one  of  the  sides  forming  the  vertex.     Show  that   if  the 

re  proportional  to  the  sides  to  which  they  are  parallel  and  if  their 

directions  are  in  a  cyclic  order  their  resultant  is  a  couple. 

10.  A  uniform  chain  of  length  /  hangs  over  a  rough  horizontal  cylinder 
of  radius  a.     Find  the  length  of  the  portions  which  hang  vertically  when 


EQUILIBRIUM  OF  RICH)   BODIES  59 

the  chain  Is  on  the  point  of  motion  under  its  own  weight,  d)  when  '/  ia 
negligible  compared  with  /,  (2)  when  it  is  nut  negligible  compared  with  /. 

11.  Two  equal  weights  arc  attached  to  the  extremities  of  a  string 

which  hangs  over  a  rough  horizontal  cylinder.  Find  the  least  amount 
by  which  either  weight  must  be  increased  in  order  to  starl  the  system  to 
move.    The  weight  of  the  string  is  negligible. 

12.  Three  cylindrical  pegs  of  equal  radius  and  roughne88  are  placed 
at  the  vertices  of  a  vertical  equilateral  triangle  the  two  lower  corners  of 
which  are  in  the  same  horizontal  line.  A  string  of  negligible  weighl  ia 
attached  to  two  weights  and  slung  over  the  pegs.  Find  the  ratio  of  the 
weights  if  they  are  on  the  point  of  motion. 

13.  A  sphere  laid  upon  a  rough  inclined  plane  of  inclination  «  is  on  the 
point  of  sliding.     Show  that  the  coefficient  of  friction  is  •  tan  a. 

14.  A  uniform  ring  of  weight  W  hangs  on  a  rough  peg.  A  bead  of 
weight  w  is  fixed  on  the  ring.     Show  that  if  the  coefficient  of  friction 

W 

between  the  ring  and  the  peg  is  greater  than— —  the  ring  will 

\  W*  +  2wW 

be  in  equilibrium  whatever  the  position  of  the  bead  with  respect  to  the 
peg- 

15.  A  uniform  rod  is  in  equilibrium  with  its  extremities  on  the  interior 
of  a  rough  vertical  hoop.     Find  the  limiting  position  of  the  rod. 

16.  A  weight  W  is  suspended  from  the  middle  of  a  cord  whose  ends  an: 
attached  to  two  rings  on  a  horizontal  pole.  If  w  be  the  weight  of  each 
ring,  fj.  the  coefficient  of  friction,  and  I  the  length  of  the  cord,  find  the 
greatest  distance  apart  between  the  rings  compatible  with  equilibrium. 


CHAPTER   IV. 
EQUILIBRIUM    OF   FLEXIBLE    CORDS. 

61.  Simplification  of  Problems.  —  The  simplest  phenome- 
non in  nature  is  the  result  of  innumerable  actions  and 
reactions.  The  consideration  of  all  the  factors  which  con- 
tribute to  any  natural  phenomenon  would  require  unlimited 
analytical  power.  Fortunately  the  factors  which  enter  into 
dynamical  problems  are  not  all  of  equal  importance.  Often 
the  influence  of  one  or  two  predominate,  so  that  the  rest  can 
be  neglected  without  an  appreciable  departure  from  the  actual 
problem.  Any  one  who  attempts  to  solve  a  physical  problem 
musl  recognize  this  fact  and"  use  it  to  advantage  by  repre- 
senting the  actual  problem  by  an  ideal  one  which  has  only 
the  important  characteristics  of  the  former.  This  was  done 
in  the  last  two  chapters  in  which  bodies  were  treated  as  single 
particles  and  rigid  bodies,  and  the  problems  were  thereby 
simplified  without  changing  their  character. 

The  same  procedure  will  be  followed  in  discussing  the 
equilibrium  of  flexible  cords,  such  as  belts,  chains,  and  ropes. 
These  bodies  will  be  represented  by  an  ideal  cord  of  negli- 
gible cross-secl  ion  and  of  perfect  flexibility.  The  solution  of 
the  idealized  problems  gives  us  a  close  enough  approxima- 
tion for  practical  purposes.  If,  however,  closer  approxima- 
1  ion  is  desired  smaller  factors,  such  as  the  effects  of  thickness 
and  imperfecl  flexibility,  may  be  taken  into  account. 

62.  Flexibility.  —  A  cord  is  said  to  be  perfectly  flexible  if  it 
offers  ii"  resistance  to  bending;  in  other  words,  in  a  perfectly 
flexible  cord  there  are  no  internal  forces  which  act  in  a 
direction  perpendicular  to  its  length. 

GO 


EQUILIBRIUM   OF   FLEXIBLE  CORDS 


61 


63.  Suspension  Bridge  Problem.  —  The  following  are  the 
important  features  of  a  suspension  bridge  which  Bhould  be 

considered  in  order  to  simplify  the  problem: 

1.  The  weights  of  the  cables  and  of  the  chains  are  small 
compared  with  that  of  the  road-bed. 

2.  The  road-bed  is  practically  horizontal. 

3.  The  distribution  of  weight  in  the  road-bed  may  be 
considered  to  be  uniform. 

We  can,  therefore,  obtain  a  sufficiently  close  approxima- 
tion if  we  consider  an  ideal  bridge  in  which  the  cable  and 
the  chains  have  no  weight  and  the  distribution  of  weight  in 
the  road-bed  is  uniform  in  the  horizontal  direction.  With 
these  simplifications  consider  the  forces  acting  upon  that 
part  of  the  cable  which  is  between  the  lowest  point  and  any 
point  P,  Fig.  41. 


Fig.  41. 


The  forces  are:  The  tensile  force  T0,  which  acts  horizon- 
tally at  0.  The  tensile  force  T,  which  acts  along  the  tangent 
to  the  curve  at  P.  The  weight  of  that  part  of  the  bridge 
which  is  between  0  and  P.  If  w  be  the  weight  per  unit 
length  of  the  road-bed  and  x  denotes  the  length  "'I'',  then 
the  third  force  becomes  w.r. 

Therefore  the  conditions  of  equilibrium  give 


Sis 

-  T0  +  T  cos  6  =  0 ; 

.-.    Fcosfl     r. 

l 

ZY  = 

-wx+Tsin  0=  0; 

:.    Tan  6  =  wx. 

(2) 

62  ANALYTICAL  MECHANICS 

It  is  evident  from  equation  (1)  that  the  horizontal  compo- 
nent of  the  tensile  force  is  constant  and  equals  T0.  Squaring 
equations  (1)  and  (2)  and  adding  we  get 

r-  =  TQ-  +  w-x\  (3) 

Thus  we  see  that  the  smallest  value  of  T  corresponds  to 
x  =  0  and  equals  T0,  while  its  greatest  value  corresponds 
to  the  greatest  value  of  x.  If  D  denotes  the  span  of  the 
bridge  then  the  greatest  value  of  T,  or  the  tensile  force  of 
the  cable  at  the  piers,  is 

In  order  to  find  the  equation  of  the  curve  which  the  cable 
assumes  we  eliminate  T  between  equations  (1)  and  (2). 
This  gives 

tan  6  =  %■  x.  (4) 

Substituting  -f-  for  tan  0  and  integrating  we  get 
ax 

1    W       »    . 

where  c  is  the  constant  of  integration. 

But  with  the  axes  we  have  chosen,  y  =  0  when  x  =  0, 
therefore  c=  0.     Thus  the  equation  of  the  curve  is 

which  is  the  equation  of  a  parabola. 

Dip  of  the  Cable.  —  Let  H  be  the  height  of  the  piers 

above  the  lowest  point  of  the  cable.     Then  for  x  =  —,y=  H, 

therefore 

H=ir0D2'  (6) 

It  is  evident  from  the  last  equation  that  the  greater  the 
ten -ion  the  less  is  the  sag. 


EQUILIBRIUM  OF  FLEXIBLE  CORDS 


63 


Problem.  A  bridge  is  supported  by  two  suspension  cables.  The 
bridge  has  a  weight  of  1.5  tons  per  horizontal  fool  and  has  a  span  of  100 
feet.    Supposing  the  dip  of  the  bridge  to  be  50  feel  find  the  values  of 

the  tensile  force  at  the  lowest  and  highesl  points  of  the  cable. 

64.  Equilibrium  of  a  Uniform  Flexible  Cord  which  is  Sus- 
pended from  Its  Ends.  —  The  problem  is  to  determine  the 
nature  of  the  curve  which  a 
perfectly  uniform  and  flexi- 
ble cable  will  assume  when 
suspended  from  two  points. 
Let  AOB,  Fig.  42,  be  the  curve. 
Consider  the  equilibrium  of 
that  part  of  the  cable  which 
is  between  the  lowest  point 
0  and  any  other  point  P. 
The  part  of  the  cable  which 
is  under  consideration  is 
acted  upon  by  the  following   three  forces: 

The  tensile  force  at  the  point  0,  T0. 

The  tensile  force  at  the  point  P,  T. 

The  weight  of  the  cable  between  the  points  0  and  /'. 
Since  the  cable  is  perfectly  flexible  T0  and  T  are  tangent  to 
the  curve.     Therefore  we  have 

ZX  =  -  T0  +  T  cos  6  =  0,     or    T  cos  0  =  T0,  I 

2  Y  =  -ws  +  T  sin  6  =  0,     or    T  sin  6  =  W8, 

where  w  is  the  weight  per  unit  length  of  the  cable  and  s  is 
the  length  of  OP. 

Squaring  equations  (1)  and  (2)  and  adding  we  obtain 

T2=TQ2+w282.  (3) 


Eliminating  T  between  equations  (1)  and  (2)  w< 

s  =  -  tan  9, 
which  is  the  intrinsic  equation  of  the  curve. 


I 


C4 


ANALYTICAL  MECHANICS 


In  order  to  express  equation  (4)  in  terms  of  rectangular 


coordinates  we  replace  tan  0  by  -g  and  obtain 
w    dx 


(5) 


But  ds2=  dx2-\-dy2,  therefore  eliminating  dx  between  this 
equation  and  equation  (5)  and  separating  the  variables 

sds 


dy  = 


Vs2  +  a2 


(6) 


and  then  integrating 


=  Vs2  +  a2  +  c, 


where  a  =  —  and  c  is  the  constant  of  integration. 


w 


Let  the  rc-axis  be  so  chosen  that  when  s  =  0,  y  =  a,  then 
c  =  0.     Therefore 


y  =  Vs2  +  a2,     or   s  =  Vy*  —  a2. 

Different  iating  equation  (7),  squaring  and  replacing 
(dx2  +  dy2)  we  have 


(7) 
f2by 


Solving  for  dx, 


dx2  +  c?i/: 


dx=  — 


_  r 


vy 


y2-a2 


ady 


- 1  Va2  -  ?/2 


(8) 


i  'Va2—y2 
where  i=  V— 1.     Integrating  equal  ion  (8)  we  get 

+  c'. 


^  ...  y 

-  =  cos  x- 
a  a 


EQUILIBRIUM  OF  FLEXIBLE  CORDS 

But  y 


a ,  when  x  = 

=  0,  therefore  c'  =  0.     Thus  w< 

1 c 
y  =  a  cos  — , 
a 

a-  * 
=  a  cosh  -, 
a 

(10) 

-fr+A 

(ID 

rp      1    102                wx\ 

2  w  v                  ' 

L2 

which  are  different  forms  of  the  equation  of  a  caU  nary. 
Discussion. — Expanding  equation  (12)  by  Maclaurin'a 

Theoremf  we  obtain 

In  the  neighborhood  of  the  lowest  point  of  the  cable  tin' 
value  of  x  is  small,  therefore  in  equation  ( 13)  we  can  neglect 
all  the  terms  which  contain  powers  of  x  higher  than  the 
second.     Thus  the  equation 

represents,  approximately,  the  curve  in  the  neighbor] d 

of  the  lowest  point.  It  will  be  observed  thai  (14)  is  the 
equation  of  a  parabola.  This  result  would  l>r  expected 
since  the  curve  is  practically  straight  in  the  neighborhood 
of  0  and  consequently  the  horizontal  distribution  of  mass 
is  very  nearly  constant,  which  is  the  important  feature  of 
the  Suspension  Bridge  problem. 

The  nature  of  those  parts  of  the  curve  which  are  removed 
from  the  lowest  point  may  be  studied  by  supposing  x  to  be 

large.     Then  since  c~~a  becomes  negligible  equation  (11 

duces  to 

•See  Appendix  Avin.  t  See  Appendix  Av. 


ANALYTICAL  MECHANICS 


The  curve,  Fig.  43,  denned 
by  equation  (15)  is  called  an 
exponential  curve.  It  has  an 
interesting  property,  namely, 
its  ordinate  is  doubled  every 
time  a  constant  value  P  is 
added  to  its  abscissa.  This 
constant  is  called  the  half -value 
period  of  the  curve.  The  value 
of  P  may  be  determined  in 
the  following  manner.  By  the 
definition  of  P  and  from  equa- 
tion (15)  we  have 

2y  =  aJ 


(16) 


Dividing  equation  (1G)  by  equation  (15)  we  get 

p 

or  P  =  a\oge  2. 

Length  of  Cable. — In  order  to  find  the  length  in  terms 
of  the  span  eliminate  y  between  equations  (7)  and  (11). 
This  gives 

a 


,ea  —e 


1     tf  1         &  , 

2  •  3  a2       2  •  3  •  4  •  5  a3 


(17) 
(18) 


where  the  right  member  of  equation  (18)  is  obtained  by 
expanding  the  right-hand  member  of  equation  (17)  by 
Maclaurin's  Theorem. 

If  I)  and  L  denote  the  span  and  the  length  of  the  cable, 
respectively,  we  have  B*=\L  when  x=\D.  Therefore 
substituting  these  values  of  s  and  as  in  equation  (18)  and 
replacing  a  by  its  value  we  obtain 


EQUILIBRIUM   OF   FLEXIBLE  CORDS  67 

When  the  cable  is  stretched  tight  T0  is  large  compared  with 
w.  Therefore  the  higher  terms  of  the  series  may  be  ae 
glected  and  equation  (19)  be  put  in  the  following  approxi- 
mate form. 

1    w2 
Hence  the  increase  in  length  due  to  sagging  is  — -         D  . 

24    T o" 
approximately. 

PROBLEMS. 

1.  A  perfectly  flexible  cord  hangs  over  two  smooth  pegs,  with  its  ends 
hanging  freely,  while  its  central  part  hangs  in  the  form  of  a  catenary.  If 
the  two  pegs  are  on  the  same  level  and  at  a  distance  D  apart,  show  that 
the  total  length  of  the  string  must  not  be  less  than  Dc,  in  order  thai 
equilibrium  shall  be  possible,  where  e  is  the  natural  logarithmic  base. 

2.  In  the  preceding  problem  show  that  the  ends  of  the  cord  will  be 
on  the  x-axis. 

3.  Supposing  that  a  telegraph  wire  cannot  sustain  more  than  the 
weight  of  one  mile  of  its  own  length,  find  the  least  and  the  greatesl  sag 
allowable  in  a  line  where  there  are  20  poles  to  the  mile. 

4.  Find  the  actual  length  of  the  wire  per  mile  of  the  line  in  the  pre- 
ceding problem. 

5.  The  width  of  a  river  is  measured  by  stretching  a  tape  over  it. 
The  middle  point  of  the  tape  touches  the  surface  of  the  water  while  the 
ends  are  at  a  height  H  from  the  surface.     If  the  tape  reads  S,  show  that 


/^- 


the  width  of  the  river  is  approximately  l 

6.  Show  that  the  cost  of  wire  and  posts  of  a  telegraph  line  is  mini- 
mum if  the  cost  of  the  posts  is  twice  that  of  the  additional  length  <>f  wire 
required  by  sagging.  The  posts  are  supposed  to  be  evenly  spaced  and 
large  in  number. 

7.  A  uniform  cable  which  weighs  loo  tons  is  suspended  between  two 
points,  500  feet  apart,  in  the  same  horizontal  line.    The  lowest  poinl  of 

the  cable  is  40  feet  below  the  points  of  support.      Find  the  .-inallot  and 
the  greatest  values  of  the  tensile  force. 

8.  In  the  preceding  problem  find  the  length  of  the  cable. 

65.  Friction  Belts.  —  The  flexible  cord  .!/>'.  Fig.  H.  ifi  in 
equilibrium  under  the  action  of  three  forces,   namely,  To 


68 


ANALYTICAL  MECHANICS 


and  T,  which  are  applied  at  the  ends  of  the  cord,  and  the 
reaction  of  the  rough  surface  of  C,  with  which  it  is  in  con- 
tact. It  is  desired  to  find  the  relation  between  T0  and  T 
when  the  cord  is  just  on  the  point  of  motion  towards  T0. 

f 
Y 


Fro.  44. 


Consider  the  equilibrium  of  an  element  of  that  part  of  the 
cord  which  is  in  contact  with  the  surface.  The  element 
is  acted  upon  by  the  following  three  forces: 

The  tensile  force  in  the  cord  to  the  right  of  the  element. 

The  tensile  force  in  the  cord  to  (he  left  of  the  element. 

The  reaction  of  the  surface. 

Lei  the  tensile  force  to  the  left  of  the  element  be  denoted 
by  T,  then  the  tensile  force  to  the  righl  may  be  denoted  by 
T+dT.  Oe  the  other  hand  if  R 'denotes  the  reaction  of 
the  surface  per  unil  Length  of  the  cord,  the  reaction  on  the 
elemenl  is  R  cfe,  where  ds  is  the  length  of  the  element.  We 
will,  as  usual,  replace  R  by  its  frictional  component  F  and 
its  normal  component  N. 


EQUILIBRIUM  OF  FLEXIBLE  CURDS 

Taking  the  axes  along  the  tangent  and  the  normal  through 
the  middle  point  of  the  element  and  applying  the  conditions 

of  equilibrium  we  obtain 

2X  -  (T  +  dT)  cos-  -  T  cos  f  -  F  (  -  ds)*  =  0, 
ZY^Nds-  T  sin  ^  -  (T  +  dT)  sin  -  =  0, 

2  2 

or  d77cos-^  +  F<is  =  0, 

and  Nds-2Tsm^--dTsm-  =  0, 

2  2 

where  dd  is  the  angle  between  the  two  tensile  forces  which 
act  at  the  ends  of  the  element.  Hut  since  the  cord  is  suit- 
posed  to  be  perfectly  flexible  the  tensile  forces  arc  tangent 
to  the  surface  of  contact.  Therefore  0  is  the  angle  between 
the  tangents,  and  consequently  the  angle  between  the  nor- 
mals, at  the  ends  of  the  element.  As  an  angle  becomes 
indefinitely  small  its  cosine  approaches  unity  and  it-  sine 
approaches  the  angle  itself,f  therefore  we  can  make  the 
substitutions 

dd     .       ,    .   dd     de 

cos  —  =  1  and  sin  —  =  — 
2  2        2 

in  the  last  two  equations,  and  obtain 

dT+Fds  =  0,  (1) 

and  Nds-Tdd+\dTdd=0. 

Neglecting  the  differential  of  the  second  order  in  equation 
(2)  and  then  eliminating  ds  between  equations  1  and  2) 
we  get 

dT         F  .„  Jn  /ox 

where  n  is  the  coefficient  of  friction.     Integrating  the  last 

*  The  negative  sign  in  F  (—ds)  indicates  the  fact  that  F  and 

measured  in  opposite  directions, 
t  See  Appendix  Avi. 


70 


ANALYTICAL  MECHANICS 


equation  and  passing  from  the  logarithmic  to  the  exponen- 
tial form,  we  have 

T  =  ce-*, 

where  c  is  the  constant  of  integration.  If  0  is  measured  from 
the  normal  to  the  surface  at  the  point  where  the  right-hand 
side  of  the  cord  leaves  contact  we  obtain  the  initial  condition, 
T  =  T0  when  0=0,  which  determines  c.  Applying  this  con- 
dition to  the  last  equation  we  have 

T  =  T0e-"e.  (4) 

Discussion.  —  Equation  (4)  gives  the  relation  between  the  values  of 
the  tensile  force  at  any  two  points  of  the  cord.  It  must  be  observed  that 
0  is  ii teasured  in  the  same  direction  as  F;  in  other  words,  opposite  the 
direction  towards  which  the  cord  is  urged  to  move.  Therefore  T  or  T0 
has  the  larger  value  according  to  whether  0  is  positive  or  negative.  As 
a  concrete  example  suppose  a  weight  W  to  be  suspended  from  the  right- 
hand  end  of  the  cord  and  to  be  held  in  equilibrium  by  a  force  F  applied  at 
the  left-hand  end.  If  F  is  just  large  enough  to  prevent  W  from  falling 
then  the  cord  will  be  on  the  point  of  moving  to  the  right,  therefore  0  is 
measured  in  the  counter-clockwise  T 
direction  and  is  positive.  In  this 
case 

F  =  We'*9. 

In  case  F  is  just  large  enough  to  start 
11'  to  move  up,  then  0  is  measured  in 
the  clockwise  direction  and  is  nega- 
tive.    Therefore 

F  =  We>*. 

The  value  of  T  drops  very  rapidly 
with  the  increase  of  0.     This  fact 
is  made  clear  by  drawing  the  graph 
of  equation  1 1 1,  Fig.  \5.    The  graph 
may  be  constructed  easily  by  making  use  of  the  half-value  period  of  the 
curve.    If  /'  denotes  the  period,  then,  by  definition,  the  ordinate  is  reduced 
-half  its  value  every  time  P  is  added  to  0.*     We  have  therefore 
lT=Toe-»i0+P). 

*  The  difference  between  this  definition  of  P  and  the  one  given  in  the  pre- 
ceding Bection  La  accounted  for  by  the  difference  in  the  signs  of  the  exponents 
in  equation    I    and  in  equation  (14)  of  the  preceding  section. 


Dividing  equation  (4)  by  the  last  equation  we  get 

2  =  &>p, 

P=ilog.2 

-1. 

us  if  6  =  nP,  then  by  equations  (4)  and  (5) 

T=T_o. 

2" 

EQUILIBRIUM  OF  FLEXIBLE  CORDS  71 


(5) 


(6) 

Therefore  taking  0.53  for  hemp  rope  on  oak  and  0  =  2t,  we  obtain 
n  =  4.76  and  2"  =  27.3.  Hence  in  this  case  T0  ia  27.3  tames  aa  greal 
as  T. 

Application  to  Belts.  —  The  tensile  force  on  one  side 
of  a  belt  which  transmits  power  is  greater  than  that  on  the 
other  side.  The  relation  between  the  tensile  forces  on  the 
two  sides  of  the  belt  is  given  by  equation  (4).  Thus  if  7\ 
denotes  the  tensile  force  on  the  driving  side  and  T2  thai  on 
the  slack  side,  then 

T2  =  Txe-»Q    or     Tx  =  Toe1*.  (4'» 

The  difference  between  7\  and  T2  is  the  effective  force  which 
drives  the  pulley.  Denoting  the  effective  force  by  /',  we 
have 

F  =  Tx  -  T2 

=  T1(l-e-"B)\  ~ 

=  T2(e"0-  1).  J 

We  have  neglected  the  cross-section  of  the  cord  in  the 
solution  of  the  foregoing  problem.  Therefore  the  results 
which  we  have  obtained  are  applicable  to  actual  problems 
only  when  the  cross-section  of  the  cord  is  negligible  com- 
pared with  that  of  the  solid  with  which  it  is  in  contact. 


72  ANALYTICAL  MECHANICS 

PROBLEMS. 

1.  A  weight  of  5  tons  is  to  be  raised  from  the  hold  of  a  ship  by  means 
of  a  rope  which  takes  3}  turns  around  the  drum  of  a  steam  windlass.  If 
p.  =  0.25  what  force  must  a  man  exert  at  the  other  end  of  the  rope? 

2.  By  pulling  with  a  force  of  200  pounds  a  man  just  keeps  from  surg- 
ing a  rope,  which  takes  2.5  turns  around  a  post.  Find  the  tensile  force 
at  the  other  end  of  the  rope,     ju  =  0.2. 

3.  A  weight  W  is  suspended  by  a  rope  which  makes  1*  turns  around 
a  clamped  pulley  and  goes  to  the  hand  of  a  workman.  If  fj.  =  0.2,  find 
the  force  the  man  has  to  apply  in  order  (a)  to  support  the  weight,  (b)  to 
raise  it. 

4.  Two  men,  each  of  whom  can  exert  a  pull  of  250  pounds,  can  sup- 
port a  weight  by  means  of  a  rope  which  takes  2  turns  around  a  post. 
On  the  other  hand,  one  of  the  men  can  support  it  alone  if  the  rope 
makes  2.5  turns.     Find  the  weight. 

6.  In  order  to  prevent  surging  a  sailor  has  to  exert  a  force  of  150 
pounds  at  the  end  of  a  hawser,  which  is  used  to  keep  the  stern  of  a  boat 
at  rest  while  the  Low  is  being  turned  by  the  engines.  Find  the  pull 
exerted  by  the  boat  upon  the  hawser  under  the  following  conditions: 

[Hint.  —  Make  use  of  equations  (5)  and  (6).] 


w»-=, 

fx  =  0.2. 

(g)    0  =  2tt, 

tx  =  0.1. 

(b)0  =  |. 

M  =  0.5. 

(h)  9-**> 

fx  =  0.4. 

(0  0  =  1 

fi  =  0.5. 

©•-*=■ 

At  =  0.5. 

(d)      0   =  7T, 

fx  =  0.4. 

(j)   0  =  3tt, 

M  =  0.3. 

(e)6  =  *f, 

/i  =  0.3. 

(k)  6  =  1^, 

Ijl  =  0.4. 

(f)  0  =  ^, 

M  =  0.2. 

(i)  e  =  7f, 

ju  =  0.5. 

6.    A  bell  has  to  transmit  an  effed 

ive  force  of  500 

pounds.     Find 

tensile  force  on  both  sides  of  the  belt, 

under  the  following  conditions 

(a)  0  =  135°, 

M  =  0.5. 

(e)    0  =  105°, 

M  =  0.2. 

(b)  6  =  135°, 

M  =  0.4. 

(f)    6  =  180°, 

At  =  0.3. 

(c)   0  =  150°, 

At  =  0.3. 

(g)  e  =  180°, 

M  =  0.5. 

id)    0        Mi.",  , 

H  =  0.5. 

(h)  d  =  195°, 

M  =  0.4. 

7.    In   the  preceding  problem    find    the  width    of   the    belt,  supposing 
the  permissible  safe  tensile  force  to  be  50  pounds  per  inch  of  its  width. 


CHAPTER   V. 

MOTION. 

FUNDAMENTAL    MAGNITUDES. 

66.  Analysis  of  Motion. — The  conception  of  motion  neces- 
sarily involves  four  ideas,  namely,  the  ideas  of 

(a)  A  body  which  moves. 

(b)  A  second  body  with  respect  to  which  it  moves. 

(c)  A  distance  which  it  covers. 

(d)  An  interval  of  time  during  which  the  distance  is 
covered. 

67.  Relativity  of  Motion.  Reference  System.  —  The  firsl 
important  inference  to  be  drawn  from  the  foregoing  analysis 
is  the  fact  that  motion  presupposes  at  least  two  bodies, 
namely,  the  body  which  is  supposed  to  move  and  the  body 
to  which  the  motion  is  referred.  The  words  "motion"  and 
"rest"  become  meaningless  when  applied  to  a  single  particle 
with  no  other  body  for  reference.  Whenever  we  think  or 
talk  about  the  motion  of  a  particle  we  refer  its  motion, 
consciously  or  unconsciously,  to  other  bodies.  The  body 
to  which  motion  is  referred  is  called  a  rqferena  system. 
The  choice  of  a  particular  body  as  a  reference  system  is  a 
question  of  convenience.  If  a  man  walk-  in  a  crowded  car 
fast  enough  to  discommode  its  occupants  he  will  be  blamed. 
not  because  he  is  moving  at  the  rate  of.  say,  20  mile-  per 
hour  with  respect  to  the  ground,  bul  because  he  i-  moving 
at  the  rate  of  4  miles  per  hour  with  respecl  to  the  car.  In 
this  case  the  car  should  be  taken  a-  the  reference  system, 
and  not  the  ground.  On  the  other  hand  if  the  man  want- 
to  leave  the  moving  car,  it  is  of  meat  importance  for  him  to 


74  ANALYTICAL  MECHANICS 

consider  the  velocity  with  which  he  is  going  to  land.  In  this 
case,  therefore,  the  surface  of  the  earth  should  be  taken  as 
the  reference  system. 

68.  Fundamental  Magnitudes.  —  The  first  two  of  the  four 
conceptions  into  which  we  analyzed  motion  are  similar; 
therefore  three  distinct  conceptions  are  associated  with 
motion.  The  first  of  these  is  the  idea  of  body,  or  of  matter; 
the  second  is  that  of  distance,  and  the  third  is  that  of  time. 
Distance  and  time  are  terms  which  are  too  familiar  to  be 
made  clearer  by  definitions,  therefore  we  will  not  attempt 
to  define  them. 

In  their  efforts  to  reduce  natural  phenomena  to  their 
simplest  terms  scientists  have  come  to  the  conclusion  that 
all  physical  phenomena  are  the  result  of  motion.  It  is 
the  main  object  of  science  to  describe  the  complicated  phe- 
nomena of  nature  in  terms  of  motion,  in  other  words,  to 
express  all  physical  magnitudes  in  terms  of  the  three  magni- 
tudes involved  in  motion.  Therefore  time,  mass,  and  length 
are  called  fundamental  magnitudes  and  all  others  derived 
magnitudes. 

69.  Fundamental  Units. — The  units  of  time,  length,  and 
mass  are  called  fundamental  units,  while  those  of  other 
magnitudes  are  called  <l<  rived  units. 

70.  The  Unit  of  Time  is  st.  \ ,,,,  part  of  the  mean  solar  day, 
and  is  called  the  second. 

71.  The  Unit  of  Length  is  the  centimeter,  which  is  -jot  Par^ 
of  the  standard  meter.  The  latter  is  the  distance  at  0°  C. 
between  two  parallel  lines  drawn  upon  a  certain  platinum- 
indium  bar  in  the  possession  of  the  French  government. 

72.  Mass.  The  choice  of  the  units  of  time  and  length  is 
comparatively  easy.  We  associate"  only  one  property  with 
each  of  these  quantities,  therefore  in  choosing  a.  unit  all  we 
have  to  do  is  to  decide  upon  its  size.  Matter,  on  the  other 
hand,  ha-  a  great  Dumber  of  properties,  such  as  volume, 
Bhape,  temperature,  weight,  mass,  elasticity,  etc.    We  com- 


MOTION  75 

pare  and  identify  different  bodies  by  means  of  these  proper- 
ties.    In  selecting  one  of  these  properties  to  represent  the 

body  in  our  study  of  motion  we  must  sec  thai  the  property 
fulfills  two  conditions:  that  it  is  intimately  related  to  motion 
and  that  it  is  constant. 

Weight  is  often  used  to  represent  a  body  in  its  motion. 
So  far  as  bodies  on  the  earth  are  concerned  weighl  La 
intimately  connected  with  motion,  but  it  is  not  constant. 
Besides,  when  bodies  are  far  from  the  earth,  weighl  does  nol 
have  a  definite  meaning.  Therefore  weight  does  Dol  satisfy 
the  foregoing  conditions.  The  property  which  serves  tin- 
purpose  best  is  known  as  mass.  It  is  intimately  connected 
with  motion  and  is  constant.*  The  nature  of  this  property 
will  be  discussed  in  the  next  chapter.  Therefore  we  will  c<  in- 
tent ourselves  by  denning  mass  as  that  property  with  which 
bodies  are  represented  in  discussions  of  their  motion. 

73.  Unit  of  Mass. — The  unit  of  mass  is  the  gram,  which 
is  yoVo  Part  °f  the.  mass  of  the  standard  kilogram.  The 
latter  is  the  mass  of  a  piece  of  platinum  in  the  possession 
of  the  French  Government. 

74.  Dimensions. — The  fundamental  magnitudes  enter  into 
the  composition  of  one  derived  magnitude  in  a  manner  dif- 
ferent from  the  way  they  enter  into  that  of  a  second.  Lengl  h 
alone  enters  into  the  composition  of  an  area,  while  velocity 
contains  both  length  and  time,  and  all  three  of  the  funda- 
mental magnitudes  combine  in  work  and  momentum.  The 
expression  which  gives  the  manner  in  which  time,  length, 
and  mass  combine  to  form  a  derived  magnitude  is  called  the 
dimensional  formula  of  that  magnitude.  Thus  the  dimen- 
sional formulae  for  area,  velocity,  and  momentum  are,  respec- 
tively, 

[A]=[L2],     [V]  =  [LT-%    and    [H]  -  [MLT^J, 

where  .1/,  L,  and  T  represent  length,  mass,  and  time.  The 
exponent  of  each  letter  is  called  the  dimension  of  the  de- 

*  C'f.  §101. 


76  ANALYTICAL  MECHANICS 

rived  magnitude  in  the  fundamental  magnitude  which  the 
letter  represents.  Thus  area  has  two  dimensions  in  length 
and  zero  dimension  in  both  time  and  mass,  while  momen- 
tum has  one  dimension  in  mass,  one  dimension  in  length, 
and  minus  one  dimension  in  time. 

75.  Homogeneity  of  Equations.  —  Magnitudes  of  different 
dimensions  can  neither  be  added  nor  subtracted.  There- 
fore in  a  true  equation  the  sum  of  the  magnitudes  of  one 
kind  which  are  on  the  left  of  the  equation  sign  equals  the 
sum  of  the  magnitudes  of  the  same  kind  which  are  on  the 
right.  When  all  the  terms  of  an  equation  have  the  same 
dimensions  the  equation  is  said  to  be  homogeneous. 

76.  Systems  of  Units.  —  The  C.G.S.  System  is  used  in 
most  of  the  civilized  countries  and  by  scientists  all  over 
the  world.  In  this  system  the  centimeter,  the  gram,  and 
the  second  are  the  fundamental  units. 

English-speaking  people  use  another  system,  known  as 
the  British  gravitational  system,  in  which  weight,  length, 
and  time  are  the  fundamental  magnitudes  and  the  pound, 
the  foot,  and  the  second  are  the  fundamental  units.  Thus 
the  unit  of  time  is  the  same  in  both  systems.  The  following 
equations  give  the  relation  between  the  centimeter  and  the 
inch  with  an  error  of  less  than  one-tenth  of  one  per  cent. 

1  in.    =  2.5-4  cms. 
1  cm.  =  0.3937  in. 

The  relation  between  the  mass  of  a  body  which  weighs  one 
pound  and  the  grain  is  given  by  the  following  equations 
with  an  error  of  less  than  one-tenth  of  one  per  cent. 

1  kg.  =  2.205  pds. 
1  ])d.  =  453.6  gms., 

where  kg.  is  the  abbreviation  for  the  kilogram,  or  1000  gms., 
while  ])d.  denote-  ihe  mass  of  a  body  which  weighs  one 
pound  in  London  and  is  often  called  pound-ma8S.     Denoting 


Fig.   16. 


MOTION  77 

the  pound  (weight)  by  its  usual  abbreviation  we  have 
2.205  lbs.  =  the  weight  of  L000  gms.* 
VELOCITY. 

77.  Displacement.  —  When  the  position  of  a  particle  with 
respect  to  a  reference  system  is  slightly  changed  it  is  Baid 
to  have  been  displaced,  and  the 
vector  s,  Fig.  4G,  which  has  its 
origin  at  the  initial  position  and 
its  terminus  at  the  final  position, 
is  called  a  displacement. 

78.  Velocity.  —  If  a  particle  un- 
dergoes equal  displacements  in 
equal  intervals  of  time,  however  • 
small  these  intervals,  it  is  said  to 
have  a  constant  velocity.  In  this 
particular  case  the  velocity  equals,  numerically,  the  distance 
covered  per  second.  When,  therefore,  a  distance  8  is  covered 
in  an  interval  of  time  t,  the  velocity  is  given  by 

s 

—v 

By  equal  displacements  are  meant  displacements  equal 
in  magnitude  and  the  same  in  direction.  Therefore  con- 
stant velocity  means  a  velocity  which  is  constant  in  direc- 
tion as  well  as  in  magnitude.  The  magnitude  «>('  velocity 
without  regard  to  its  direction  is  called  speed. 

In  general,  bodies  not  only  cover  unequal  distances  in 
equal  intervals  of  time,  but  also  change  their  directions  of 
motion.  Therefore  we  need  a  definition  of  velocity  like  the 
following,  which  is  perfectly  general. 

The  velocity  of  a  particle  at  any  /><>in(  of  its  path  equals, 
in  magnitude,  the  time  rod  <il  which  it  describes  that  pari  of 
the  path  which  is  in  the  immediate  neighborhood  of  tht  point 
and  has  the  direction  of  lh<  tangent  at  that  /mint. 

*  For  t lie  relation  between  mass  and  weight    a   p    109. 


78 


AXALYTH  AL  MECHANICS 


FlO.    !7 


In  order  to  expiv»  this  definition  of  velocity  in  analytical 
language,  consider  a  particle  describing  a  curved  path  with 
a  changing  speed.  The  most  natural  way  of  determining 
the  s] )<■<•<  1  at  a  point  P,  Fig.  47,  is  to  observe  the  interval 
of  time  which  the  particle  takes  to  pass  two  points,  Pi  and 
7J.2,  which  arc  equidistant  from  P,  y 
then  to  divide  the  distance  P1P2  by 
thai  interval  of  time.  This  gives 
the  average  speed  from  Pi  to  P2, 
which  may  or  may  not  equal  the 
actual  speed  at  P.  If,  however, 
we  take  the  points  Pi  and  P2  nearer 
to  P  we  obtain  an  average  speed 
which  is,  in  general,  nearer  the 

speed  at  P,  because  there  is  less     

chance  for  large  variations.     If  we      ° 

take  Pi  and  P2  nearer  and  nearer 

the  average  speed  approaches  more  and  more  to  the  value  at 

P  P« 

P.     Therefore  the  limiting  value  of  the  ratio  — -- 2-  is  the 

speed  at  P.     In  other  words 

„=!=*.  a) 

is  the  analytical  definition  of  speed.  Therefore  the  velocity 
is  a  vector  which  has  s  for  its  magnitude  and  which  is  tan- 
gent to  the  path  at  the  point  considered,  that  is, 

v  =  s.  (I') 

*  The  Differential  Calculus  was  invented  by  Newton  and  Leibnitz  inde- 
pendently. Newton  adopted  a  notation  in  which  the  derivative  of  a  variable 
8  with  respect  to  another  variable  is  denoted  by  i.  This  notation  is  not  con- 
venient when  derivatives  arc  taken  with  respect  to  several  variables.  The 
notation  introduced  by  Leibnitz  is  more  convenient  and  is  the  notation  which 
ifl  generally  adopted.  Newton's  notation,  however,  is  often  Used  to  denote 
differentiation   with    respect    to   time.     On  account  of  the  compactness  of  ,s 

Compared   with  -jj,  we  will   denote  differentiations  with   respect  to   time  by 

Newton's  notation  whenever  compactness  of  expression  is  desired. 


MOTION 


79.  Dimensions  and  Units  of  Velocity.  —  The  dimensions 

of  velocity  are  [LT~~X\.    The  C.G.S.  unit  of  velocity  is  the 

cm, 
— ' .    The  British  unit  of  velocity  is 
sec. 

■>L  . 
sec. 

80.  Rectangular  Components  of  Velocity.  —  Let  v,  Fig.  48, 


centimeter  per  second, 

the  foot  per  second, 


A 


Fig.   is. 

denote  the  velocity  at  P,  then  the  magnitude  of  its  compo- 
nent along  the  z-axis  is 

vt  =  V  cos  0 
ds 
=  dtC°S 


_  ds  cos  e 
dt 
dx      - 

=  dt=x- 


Similarly 
and 


dy 
dt 

dz 
dt 


(ID 


Equations  (II)  state  that  the  component  of  the  velocity  of 

a  particle  along  any  line  equals  the  velocity  of  the  | 

tion  of  the  particle  upon  that  line,  in  other  words,  the  ve- 


80 


ANALYTICAL  MECHANICS 


locity  along  any  direction  equals  the  rate  at  which  distance 
is  covered  along  that  direction. 

The   velocity  and  its   components  evidently  fulfill  the 

relation  

v=Vvx*+vvi+v,2.  (Ill) 

When,  as  in  the  case  of  Fig.  48,  the  particle  moves  in  the 
jcy-plane,  z  =  0,  therefore 


The  direction  of  v,  in  this  case,  is  given  by 

tan0  =  H-, 
x 

where  0  is  the  angle  v  makes  with  the  x-axis. 


(HI') 


(IV) 


ILLUSTRATIVE  EXAMPLE. 

Find  the  path,  the  velocity,  and  the  components  of  the  velocity  of  a, 
particle  which  moves  so  that  its  position  at  any  instant  is  given  by  the 
following  equations: 

x  =  at,  (a) 

y=-hgt*.  (b) 

Eliminating  t  between  (a)  and  (b),  we  obtain 

2a2 

for  the  equation  of  the  path,  therefore  the  path  is  a  parabola,  Fig.  49. 

To  find  the  component -velocities  we  differentiate  (a)  and  (b)  with 
respect  to  the  time.    This  gives  Y 

x  =  a, 

y  -  -</<• 


.-.    v  =  \/«2  +  gt2. 

Discussion.  —  The  horizontal  compo- 
nent of  the  velocity  IS  directed  tn  the  right 
and  is  constant,  while  the  vertical  com- 
ponent is  directed  downwards  and  increases 
at  a  constant  rate. 

We  will  Bee   later  that   these  ('([nations  Il,;-  49. 

represent  the  motion  of  a  body  which  is  projected  horizontally  from  an 
elevated  position. 


MOTION 


81 


PROBLEMS. 

1.   Find  the  path  and  the  velocity  of  a  particle  which  moves  bo  that 
its  position  at  any  instant  is  ijiven  by  the  following  pairs  of  equal  ii 


(a)  x  =  at, 

V  =  ht. 

(b)  x  =  at, 

y  =  at-  \  gt°- 

(c)  X  =  at, 

y  =  b  cos  ut. 

(d)  X  =  a  sin  ut, 

V  =  bt. 

(e)  x  =  a  sin  ut, 

y  =  a  cos  ut. 

(f)  x  =  a  sin  ut, 

y  =  b  sin  u& 

(g)  x  =  aekt, 

v/  =  ae~kt. 

ve  the  relation  v  =  Vx2  +  y 

2  +  i2. 

81.  Radial  and  Transverse  Components  of  Velocity.  —  The 
magnitude  of  the  velocity  along  the  radius  vector  is,  accord- 
ing to  the  results  of  the  preceding  section, 

a) 


Vr 


ill 


The  expression  for  the  velocity  at  right  angles  to  r  is  ob- 
tained by  considering  the  motion  of  the  projection  of  tin- 
particle  along  a  perpendicular  y 
tor.  When  the  particle  moves 
through  ds,  its  projection 
moves  through  r  dd,  Fig.  50, 
therefore  the  required  velocity 
is 

rdd 


dt 

do       -n 
r—  =rd. 
dt 


(2) 


The  components  vr  and  vp  may  be  expressed  in  terms  of  x 
and  y  by  differentiating  the  equations  of  transformation 

r2=a;2+ya  (3) 

A  (4) 


and 


6  =  tan" 


82 


ANALYTICAL  MECHANICS 


with  respect  to  the  time.     Differentiating  (3)  we  obtain 
dr 
V'=dt 
_x  dx     y  dy 
r  dt      r  dt 
=  i  cos  0  +  ?/ sin  0.  (5) 

Differentiating  (4)  we  get 

dd 

v  =  r  — 

p        dt 

_rxy-yx 

x2  +  y2 

=  ycosd-x  sin  0.  (6) 

These  components  satisfy  the  relation 

v  =  Vr2+  r-b\  (7) 


ILLUSTRATIVE   EXAMPLE. 
A  particle  describes  the  motion  defined  by  the  equations 

x  =  a  cos  kt,  (a) 

and  y  =  a  sin  kt.  (b) 

Find  the  equation  of  the  path,  the  velocity  at  any  instant,  and  the  com- 
ponents of  the  latter. 

Squaring  and  adding  (a)  and  (b)  we  eliminate  t  and  obtain 
X2  -f  if  =  o*  Y 


for  the  equation  of  the  path. 
Differentiating  (a),  we  have 

.       dx 
X  =  dJ 

=  —ha  si  i 

kt 

=  -ky. 

Differentiating  (b), 

we  obtain 

.      dy 

=  ka  ens  kt 
=  fee. 


MOTION 


83 


Therefore 


v  =  Vx-  +  y- 


=  k  Vx-  +  if 
=  ka. 


Thus  the  particle  describes  a  circle  with  a  constant  speed  ka.    The  direc- 
tion of  the  velocity  at  any  instant  is  given  by  the  relation 

tan  6  =  I 


The  components  vr  and  vp  may  be  obtained  at  once  by  remembering,  (1) 
that  the  radius  vector  is  constant:  e.g.,  f  =  0,  (2)  that  it  is  always  normal 
to  the  path:  e.g.,  rdd  =  ds.     Therefore 

dr      n 

v*  =  Tt  =  °» 
at 


and 


dd      ds 
Tdt=dt=V 


ka. 


82.    Velocity  of  a   Particle   Relative  to  Another  Particle   in 

Motion. — Consider  the  motion  of  a  particle  Pi,  Fig.  52,  with 

Y 


o  x. 


Fia.  52. 


respect  to  a  particle  P2,  when  both  arc  in  motion  relative 
to  the  system  of  axes  XOY. 

Let  the  system  of  axes  A"'/M"  have  /',  for  it-  origin  and 
move  with  its  axes  parallel  to  those  of  the  system  XOY. 
Further  let  (x1}  yx)  and  >x:.  //v  be  the  positions,  and  v,  and 
v2  the  velocities  of  1\  and  Ps  with  respeel  to  XOY.     Thru 


84 


ANALYTICAL  MECHANICS 


if  (x',  y')  denotes  the  position  and  v'  the  velocity  of  Pi  with 
respect  to  X'P-iY',  we  get 

x'  =  Xi  —  x2, 

y'  =  yi  -  yi- 

Differentiating  the  last  two  equations  with  respect  to  the 
time 


Therefore 


x'  =  Xi  -  x2, 

\i  =  i/i  -  i/2. 

v'  =  x'  +  y' 
=  (xi  +  yi' 
=  Vi  -  v2. 


,X2+  \T2 


(V) 


Equation  (V)  states  that  the  velocity  of  a  particle  with 
respect  to  another  particle  is  obtained  by  subtracting  the 
velocity  of  the  first  from  that  of  the  second. 

ILLUSTRATIVE   EXAMPLE. 

Two  particles  move  in  the  circumference  of  a  circle  with  constant 
speeds  of  v  and  2  v.     Find  their  relative  velocities. 

Lei  the  slower  one  be  chosen  as  the  reference  particle,  and  let  the  angle 
P2OP1,  Fig.  53,  be  denoted  by  6.    Then  the  velocity  of  Pi  relative  to  P2  is 

v/  =  Vi  -  V2. 

But  vi  =  2v  and  v2  =  v,  therefore 
v' 


Fig.  53. 

Discussion.  —  Whenever  Pi  passes  P2  the  value  of  6  is  a  multiple  of 
2  7r,  therefore  cos  6  =  1  and  v'  =  v.     When  the  particles  occupy  the  ends 


MOTION  85 

of  a  diameter  cos  6  =  -  1,  therefore  Vi  -  3  v.    When  they  are  Beparated 
by  an  angle  which  is  an  odd  multiple  of  -  ,  cos  0  =  U;  therefore  d        \  5 


PROBLEMS. 

1.  An  automobile  is  moving  at  the  rate  of  30  miles  an  hour  in  a  direc- 
tion at  right  angles  to  a  train  which  is  making  40  miles  an  hour;  find  tin- 
velocity  of  the  automobile  with  respect  to  the  train. 

2.  Two  trains  pass  each  other  on  parallel  tracks,  in  opposite  directions. 
A  passenger  in  one  of  the  trains  observes  thai  it  takes  the  other  train  4 
seconds  to  pass  him.  What  is  the  length  of  the  other  train  it"  the  veloci- 
ties of  the  two  trains  are  50  and  40  miles  per  hour? 

3.  A  man  of  height  h  walks  on  a  level  street  away  from  an  electric 
lamp  of  height  //.  If  the  velocity  of  the  man  is  v,  find  the  velocity  of  the 
end  of  his  shadow  (a)  with  respect  to  the  ground  and  (l>)  with  respect  to 
the  man. 

4.  Two  particles  move,  in  opposite  directions,  on  the  circumference  of 
the  same  circle  with  the  same  constant  speed.  Find  an  expression  for 
their  relative  velocity  and  see  what  this  expression  becomes  at  BpeciaJ 
positions  of  the  particles. 

5.  A  train  is  moving  due  north  at  the  rate  of  50  miles  an  hour.  The 
wind  is  blowing  from  the  southeast  with  a  velocity  of  20  miles  an  hour. 
Find  the  apparent  direction  and  magnitude  of  the  wind  to  a  man  on  tin- 
train. 

6.  The  wind  seems  to  blow  from  the  north  to  an  automobile  party 
traveling  westward  at  the  rate  of  15  miles  an  hour.  On  doubling  the 
speed  of  the  automobile  the  wind  appears  to  come  from  the  northwest. 
Find  the  actual  direction  and  magnitude  of  the  velocity  of  the  wind. 

7.  Find  the  velocity  of  a  particle  moving  on  the  circumference  of  a 
circle  with  uniform  speed  relative  to  another  particle  moving  with  equal 
speed  in  a  diameter  of  the  circle. 

8.  Express  the  speed  of  a  mile  a  minute  in  the  C.G.S.  units. 

9.  Express  the  C.G.S.  unit  of  velocity  in  miles  per  hour. 

10.  Trove  that  i2  +  if  =  r2  +  r-0-. 

11.  Trove  analytically  that 

vz  =  vrcoa6  —  Vps'md, 
vy  =  i;r  sin  0  +  vp  i 

12.  Trove  graphically  that 

vT  =  VxCohO  +  pysin  9, 
vp  =  rvcos0  —  /'iSintf. 


86 


ANALYTICAL  MECHANICS 


A N C!  U  LAB    VELOCITY. 

83.  Angular  Displacement.  —  When  the  motion  of  a  particle 
is  referred  to  an  axis,  then  the  angle  which  the  axial  plane, 
i.e.,  the  plane  determined  by  the  particle  and  the  axis, 
describes,  is  called  an  angular  displacement.  Angular  dis- 
place! nent  is  a  vector  magnitude 
which  is  represented  by  a  vector 
drawn  along  the  axis;  as  in  the 
case  of  the  vector  representation 
of  a  torque.  The  directional  rela- 
tions are  the  same;  that  is,  the 
vector  points  towards  the  observer 
and  is  considered  as  positive  when 
the  rotation  is  counter-clockwise. 
It  points  away  from  the  observer 
and  is  negative  when  the  rotation 
is  clockwise. 

The  relation  between  the  linear 
displacement  of  a  particle  and  its 
angular  displacement  about  an  axis  may  be  found  from  a 
consideration  of  Fig.  54: 

r 
_  ds  cos  <j> 
r 

where  ds  is  the  linear  displacement  of  the  particle  P,  dd  is 
the  corresponding  displacement  about  an  axis  through  the 
point  0  perpendicular  to  the  plane  of  the  paper,  and  4>  is 
the  angle  ds  makes  with  the  normal  to  the  axial  plane. 

When  r  is  constant  <f>  is  zero,  and  the  particle  describes  a 
circle,  in  which  case  the  last  equation  becomes 


Fig. 


dd  = 


ds 


or     6 


MOTION  87 

84.  Unit  Angle.  —  In  the  last  equation  0=  1  when  s=r; 
therefore  the  angle  which  is  subtended  at  the  center  of  a 
circle  by  an  arc  equal  to  the  radius  is  the  unit  of  angle. 
This  unit  is  called  the  radian.  Angles  and  angular  dis- 
placements have  no  dimensions.     Why? 

85.  Angular  Velocity. — The  conception  of  angular  velo<  ity 
is  similar  to  that  of  linear  velocity.  It  is  the  time  rate  al 
which  the  axial  plane  sweeps  over  an  angle.  When  con- 
stant it  is  numerically  equal  to  the  angle  swept  over  per 
second.  If  we  denote  the  angular  velocity  by  o>  its  magni- 
tude is  defined  by 

Angular  velocity  is  a  vector  quantity  which  is  represented 
by  a  vector  drawn  along  the  axis  of  rotation.  The  vector 
points  towards  the  observer  when  the  rotation  is  counter- 
clockwise, and  away  from  the  observer  when  it  is  clockwise. 
The  angular  velocity  is  said  to  be  positive  in  the  firsl  case 
and  negative  in  the  second  case.  Angular  velocity  has  the 
dimensions  of  the  reciprocal  of  time. 

[0]-[r-*]. 

The  unit  of  angular  velocity  is  the  radian  per  second, — '-- 

8^  C. 

The  relation  between  the  linear  and  the  angular  velocities 

of  a  particle  may  be  obtained  from  equations  I  \  1  I  and    I  . 


"=It 


de 

dt 

ds  COS  <;, 


dt 


=«,  mi 


88 


ANALYTICAL  MECHANICS 


where  vp  is  the  component  of  the  linear  velocity  in  a  direction 
perpendicular  to  the  axial  plane. 

ILLUSTRATIVE   EXAMPLE. 

A  particle  describes  a  circle  of  radius  a  with  a  constant  speed  v.    Find 
its  angular  velocity  relative  to  an 
axis  through  a  point  on  the  cir- 
cumference  and  perpendicular  to 
the  plane  of  the  circle. 

Lei  P  (Fig.  55)  be  the  position 
of  the  particle  and  0  the  point  at 
which  the  axis  of  reference  inter- 
sects the  circle.  Move  the  par- 
ticle from  P  to  P'  and  denote  the 
linear  and  angular  displacements 
by  ds  and  (19  respectively.  Then 
the  angle  subtended  by  PP'  at 
0  is  one-half  that  subtended  at  C.     Hence 


dd  =  $d<i> 


1  ds 

2  a  ' 

de 

dt 

2  a  dt 
v 
2a 


Thus  the  angular  velocity  about  O  is  independent  of  the  position  of  the 
particle  and  equals  one-half  the  angular  velocity  aboul  the  center. 


PROBLEMS. 

1.  The  radius  of  the  earth  is  4000  miles  and  thai  of  its  orbit  93  million 

miles.     Compare  the  angular  velocities  of  a  point  on  the  equator  with 
respect  to  the  bud  al  midday  and  midnight. 

2.  In  wli.-it  latitude  is  a  bullet,  which  is  projected  east  with  a  velocity 
of  1320  feel  per  Becond,  at  rest  relatively  to  the  earth's  axis;  the  radius 
being  taken  as  WOO  miles? 


MOTION  89 

3.  A  belt  passes  over  a  pulley  which  has  a  diameter  of  30  inch 
which  makes  200  revolutions  per  minute.     Find  the  linear  Bpeed  of  tin- 
belt  and  the  angular  .speed  of  the  pulley. 

4.  The  wheels  of  a  bicycle,  which  are  75  em.  in  diameter,  mal 
revolutions  in  65  minutes.     Bind  the  Bpeed  of  the  rider;    the  angular 
speed  of  the  wheels  about  their  axles;  the  relative  velocity  of  the  highest 
point  of  each  wheel  with  respect  to  the  center. 

5.  A  point  moves  with  a  constant  velocity  v.  Find  its  angular  veloc- 
ity about  a  fixed  point  whose  distance  from  the  path  is  ". 

6.  A  railroad  runs  due  west  in  latitude  X.  Find  the  velocity  of  the 
train  if  it  always  keeps  the  sun  directly  south  of  it. 

7.  Find  the  expression  for  the  angular  velocity  of  any  point  on  the 
rim  of  a  wheel  of  radius  a,  moving  with  a  velocity  v,-  the  wheel  is  supposed 
to  be  rolling  without  slipping.  Discuss  the  values  of  the  velocity  for 
special  points. 

8.  In  the  preceding  problem  find  the  relative  velocity  of  any  point  on 
the  rim  with  respect  to  the  center  of  the  wheel,  and  the  velocity  of  the 
center  with  respect  to  the  point  of  contact  with  the  ground. 

9.  The  end  of  a  vector  describes  a  circle  at  a  constant  rate.  If  the 
origin  is  outside  the  circle  find  the  velocity  along  and  at  right  an| 

the  vector.     Discuss  the  values  for  interesting  special  positions. 

10.  In  the  preceding  problem  derive  an  expression  for  the  angular 
velocity  of  the  vector  and  discuss  it. 

ACCELERATION 

86.  Acceleration.-  -When  the  velocity  of  a  particle  changes 

it  is  said  to  have  an  acceleration.  The  change  may  lie  in 
the  magnitude  of  the  velocity,  in  the  direction,  or  in  both; 
further  it  maybe  positive  or  negative.  Therefore  the  term 
acceleration  includes  retardation  as  well  :is  increase  in  ve- 
locity.    Retardation  is  negative  acceleration. 

If  the  particle  moves  in  a  straight  path  with  :i  velocity 
which  increases  or  diminishes  at  a  constant  rate  it-  accel- 
eration equals,  numerically,  the  change  in  the  velocity  per 
second  and  is  said  to  be  constant  : 

f         V2  ~  Vl 

f_  t         ' 

where  f  is  the  acceleration  and  v,  and  v-_.  are  the  velocities 
at  the  beginning  and  at  the  end  of  the  interval  of  time  t. 


90  ANALYTICAL  MECHANICS 

Since  vi  and  v2  are  in  the  same  line,  their  difference  will  be  a 
vector  in  the  same  direction.  Therefore  in  this  particular 
case  the  acceleration  is  constant  not  only  in  magnitude  but 
also  in  direction. 

The  following  definition  of  acceleration  is  general  and 
holds  true  whatever  the  manner  in  which  the  velocity 
changes. 

The  magnitude  of  the  acceleration  of  a  particle  at  any  point 
<>f  its  path  equals  the  time  rate  at 
which  its  velocity  changes   at   the 
instant  it  occupies  that  point. 

The  analytical  expression  for 
this  definition  may  be  obtained 
by  a  reasoning  similar  to  thai 
employed  in  deriving  the  analyt- 
ical definition  of  velocity.  Sup- 
pose    it    is    required   to    find   the 

acceleration  at  P  (Fig.  56).  Let  vi  and  vo  denote  the  veloc- 
ities at   two  neighboring  points  Pj  and  P2.     Then  the  ratio 

gives  the  average  rate  at  which  the  velocity  changes  during 
the  interval  of  time  t,  which  it  takes  the  particle  to  move 
from  Pi  to  Pz.  Therefore  f  is  the  average  acceleration  for 
that  interval  of  time.  In  general  this  average  acceleration 
will  not  be  the  same  as  the  acceleration  at  P.  But  by  taking 
/',  and  I',  nearer  and  nearer  to  P  the  difference  between  the 
average  acceleration  and  the  required  acceleration  may  be 
made  as  small  as  desired.  Therefore  at  the  limit  when  Ph  P, 
and  /'.  become  successive  positions  of  the  particle,  the  aver- 
age acceleration  becomes  identical  with   the  acceleration  at 

/'.  and  the  lasl  equation  takes  the  form 

f-S-*.  (VIII) 


MOTION  91 

It  must  be  remembered  that  dv  is  the  vector  difference  of 
the  velocities  at  the  beginning  and  at  the  end  of  the  interval 
of  time  dt;  therefore  f  is  a  vector  magnitude  with  a  direction 
which  is,  in  general,  different  from  that  of  the  velocity. 

87.  Dimensions  and  Units  of  Acceleration.  The  dimen- 
sions of  acceleration  are  [LT~2].  The  unit  of  acceleration 
is  a  unit  change  in  the  velocity  per  second.     Therefore  the 

C.G.S.  unit  is   — or  '-.     Thus  if  the  velocity  of  a 

sec.  sec- 

cm 
particle  increases  by  an  amount  of  one      —  during  cadi 

sec. 

second  it  has  a  unit  acceleration.      The  engineering  unit  of 

ft 
acceleration  is  the  foot  per  sec.  per  sec,  -— '—• 

sec2 

PROBLEMS. 

1.  Express  the  engineering  unit  of  acceleration  in  terms  of  the  <  '.<  '■..s;. 
unit. 

2.  Taking  the  value  of  the  gravitational  acceleration  to  be  980 

find  its  value  in  — \  and  - — — ■ 
sec.2  hr.- 

3.  A  train  moving  at  the  rate  of  30  kilometers  per  hour  is  brought 

to  rest  in  two  minutes.     Find  the  average  acceleration  and  express  it  in 

ft.        ,  km. 

and  - — -  • 

sec*    sec.2         hr.- 

88.  Components  of  Acceleration  along  Rectangular  Axes. 
Suppose  a  particle  to  describe  a  path  in  the  ./•//-plane.     Then 
if  vi  and  v2  be  the  velocities  at  two  neighboring  points,  we 

can  write 

dv  =  y>  —  vi 

=  (x2+y-.)-(xi+y.) 
=  dx-f  dy. 
,_dv_dxdy 
'"•  dt  "  dt  +  dt  ' 

But  since  f  =  fx  +  f*, 

dx      dy 
dt      dt 


92  ANALYTICAL   MECHANICS 

The  last  equation  cannot  be  true  unless 

Tl      dt 
and  f„=f- 

Therefore  the  component  of  the  acceleration  along  a  fixed 
line  equals  the  time  rate  of  change  of  the  component  of  the 
velocity  along  that  line. 

It  follows  from  the  last  two  equations  that: 


f  _dx  _  d^x  _  ■■ 

Jx~  dt~  dt  ~X' 

_dy_d*yy 

Jv~  dt~  dt       V' 


(IX) 


The  magnitude  and  the  direction  of  the  acceleration  are 
given  by  the  following  equations: 

/=vpTF,  (x) 

tan0=^  (XI) 

x 

where  9  is  the  angle  f  makes  with  the  ar-axis. 

89.  Tangential  and  Normal  Components  of  Acceleration.  — 
The  tangential  component  of  the  acceleration  at  P  (Fig.  57) 
equals  the  rate  at  which  the  velocity  increases  along  the  di- 
rection of  the  tangent  at  P.  In  order  to  find  this  rate  we 
consider  the  velocities  at  two  neighboring  points  1\  and  P2. 
Let  vi  and  v2  be  the  velocities  at  these  points  and  ei  and  e2 
the  angles  which  vi  and  v2  make  with  the  tangent  at  P. 
Then  the  change  in  the  velocity  along  the  tangent  at  P, 
while  the  particle  moves  from  Pi  to  P2,  is 
Vi  cos  f.j  —  r,  cos  ei. 

Dividing  this  by  the  corresponding  interval  of  time  we  ob- 
tain the  average  rate  at  which  the  velocity  increases  from 


MOTION 


Pi  to  P2  along  the  tangent  at  P.     Therefore  the  average 
tangential  acceleration  is 

7-  _  V-2  COS  e2  —  Vi  COS  ei 

3t~         t 

This  average  approaches  the  actual  tangential  acceleration 
at  P  as  Pi  and  P2  are  made  to  approach  P  as  a  limit.     But 

N 


Fig.  57. 


as  these  points  approach  P  the  angles  ei  and  €2  approach 
zero  as  a  limit  and  their  cosines  approach  unity.  Therefore 
the  tangential  acceleration  at  P  is 


fT  =  limit 


3,1*7*] 


dv 
dt  ~~ 
d-s 
dt2 


s. 


By  similar  reasoning  we  obtain 

r.,  sin  e-j  -  r,  sin  ei 


/»  = 


for  the  average  normal  acceleration  between  Pi  and  P§.    The 

actual  normal  acceleration  at  P  is  the  limiting  value  of  this 
expression  as  Pi  and  P2  approach  P.      But  as  these  points 

approach  P,  ^  and  r_.  approach  v,  the  velocity  al  P,  while 


94  ANALYTICAL  MECHANICS 

sin  ei  and  sin  *>  approach  ci  and  €2,*  respectively.     Therefore 
the  normal  acceleration  at  P  is 

/„  =  limit 


[mitr_,ii±^] 

<=o    L  t     A 

imit    —  v  -\ 
t*o    L          LA 


limit 
dd 

--•a—*.  (xii) 

where  0  =  ci+€2  is  the  total  change  in  the  direction  of  the 
velocity  in  going  from  Pi  to  P2.  Since  the  direction  of  the 
velocity  coincides  with  that  of  the  tangent,  0  is  the  rate  at 
which  the  directions  of  the  tangent  and  the  normal  change. 
But  the  rate  at  which  the  normal  changes  its  direction  equals 
the  angular  velocity  of  the  particle  about  the  center  of 
curvature.  Therefore  if  p  denotes  the  radius  of  curvature 
at  P,   we  have 

i-v  ,      ***** 

and  /.  =  -  ~  (XIII) 

The  negative  sign  in  (XIII)  shows  that/n  and  p  are  measured 
in  opposite  directions.  Since  p  is  measured  from  the  center 
of  curvature,  /„  must  be  directed  towards  the  center  of  cur- 
vature. Therefore  the  total  acceleration  is  always  directed 
towards  the  concave  side  of  the  path. 

The  following  are  the  principal  results  obtained  in  this 
section  and  the  conclusions  to  be  drawn  from  them. 

(a)  The  magnitude  of  the  tangential  acceleration  is  ft; 

fr    =    ft. 

(b)  The  normal  acceleration  is  directed  towards  the  center 

v2 
of  curvature  and  lias  —  for  its  magnitude; 

Jn  p 

*  Bee  Appendix  An. 


MOTION  95 


(c)  The  magnitude  of  the  total  acceleration  is  given  by 


r 


the  relation  /=  V  v-  +  -• 

(d)  The  total  acceleration  is  directed  towards  the  concave 
side  of  the  path  and  makes  an  angle  with  the  tangent  which 

is  defined  by 

fn         v2 
toaj-j.  —  -. 

(e)  When  the  path  is  straight,  that  is,  when  p  =  oo ,  the 
normal  acceleration  is  nil;  therefore  in  this  case  the  total 
acceleration  is  identical  with  the  tangential  acceleration. 

(/)  When  the  path  is  circular  and  the  speed  constant, 
then  p  =  r,  the  radius  of  the  circle,  and  v  =  0;  therefore 

/  =  /»=--• 

J      J»  r 

90.  Radial  and  Transverse  Components  of  Acceleration.  — 
Let  P  (Fig.  58)  be  any  point  of  the  path  at  which  the 
acceleration  of  the  particle  is  to  be  considered.  Take  two 
neighboring  points  Pi  and  P2,  and  let  vi  and  v2  be  the  veloci- 
ties at  these  points.  Then  the  change  in  the  radial  velocity 
in  going  from  Pi  to  P2  is  obtained  by  subtracting  the  radial 
component  of  Vi  from  that  of  v2.  Replace  Vi  and  vs  by 
their  components  along  and  at  right  angles  to  r{  and  r-:, 
respectively,  and  denote  these  components  by  vr;,  v  and 
vr,  v/):;  then  it  will  be  seen  from  the  figure  that 

(iV,  cos  e2  -  vP2  sin  e2)  -  Ov,  cos  €i  +  vPl  sin  d) 

is  the  total  change  in  the  radial  velocity.  Therefore  the 
radial  component  of  the  acceleration  is 

f  -Y    -t  [X  cos  62  -  vp,  sin  ei  —  vr,  cos  ct  -  vPl  sin  f,1 

where  t  is  the  time  taken  by  the  particle  to  go  from  Pi  to  /'.■. 
But  as  the  points  Pi  and  P  approach  !'  as  a  limit,  the  follow- 
ing substitutions  become  permissible. 


96 


ANALYTICAL  MECHANICS 


COS  €i  =   COS  62 


Sill  ei  =  d,      sin  e2  =  62. 


tv,  -  vry  =  dvr,     vPl  =  vPl  =  vp,     ei  +  «2  =  de. 
Alaking  these  substitutions  in  the  expression  for/r,  we  obtain 
/,-  limit  [('■--^ -->(«■  +  «»)] 


dVr 


de 


dt       pdt 


dt2      r\dt), 
where  0  is  the  angle  r  makes  with  the  a>axis. 


Fig.  .58. 


(XIV) 


By  similar  reasoning  we  obtain  the  following  expressions 
for  the  transverse  acceleration,  that  is,  the  component  of 
the  acceleration  along  a  perpendicular  to  the  radius  vector. 

*  See  Appendix  Avi. 


MOTION 


J  =  limit  \Vr* Sm  ** +  Vpt  C°S  e2  ~  ^  ~  'Vl  Sin  €l  +  ^"  C0S  6l^l 
=  limit  pvfa  +  ^+fa-^j 


r 

dvp 

^  dt 

u. 

dr 

dt 

+  S™ 

1 

r 

d 
dt 

:»-2«), 

x\ 


where  co  is  the  angular  velocity  of  the  radius  vector. 


PROBLEMS. 

1.  A  particle  describes  the  parabola  y  =  2  p x  so  that  its  velocity 
along  the  x-axis  is  constant  and  equals  //.  Find  the  total  velocity  and  the 
acceleration. 

2.  Discuss  the  motions  defined  by  the  following  equations  deriving 
the  expressions  for  the  path,  velocity,  acceleration,  and  the  various  com- 
ponents of  the  last  two : 

(a)  x  =  at,        y  =  b  ^1.  (d)  .r  =  at,  y  =  i- 

(b)  x  =  at,        y  =  bt  —  \gt-.  (e)  x  =  at,  y  =  bean  tat. 

(c)  x  =  aekt,     y  =  bekt.  (f)  x  =  a  cos  cot,     y  =  lit. 

3.  Express  in  terms  of  t  the  velocity  and  the  acceleration  of  a  particle 

which  moves  so  that  x  =  ay  and  y  =  ax. 

4.  A  particle  describes  a  circle  of  radius  a  with  a  constant  Bpeed  V. 
Find  fx,  /„,  f2,  fp,  fr,  fn,  fT,  and  /. 

6.   Work  out  the  preceding  problem  graphically. 

91.  Angular  Acceleration. — Angular  acceleration  is  the  time 
rate  at  which  angular  velocity  changes.  Therefore,  denoting 
it  by  7,  we  have 

doj 


y=dt 

_  drd 
~  dt- 


(XVI 


If  the  angular  velocity  of  a  body  increases  uniformly 

'-  in  one  second  the  body  is  Baid  to  have  a  unit  angular 

1  sec. 


98  ANALYTICAL  MECHANICS 

acceleration.     Therefore  the  unit  is  the  V,.    The  dimen- 

sec- 


sions  of  angular  acceleration  are  given  by  [T~ 


ILLUSTRATIVE  EXAMPLE. 
A  particle  moves  so  that  the  coordinates  of  its  position  at  any  instant 
are  given  by  the  equations 

x  =  a  cos  kt, 

y  =  a  sin  kt. 

Find  the  acceleration  and  its  components. 

In  a  previous  illustrative  example,  p.  82,  it  was  shown  that  these 
equations  represent  uniform  circular  motion,  with  the  following  data: 


V    = 

ka, 

CO 

=  k, 

Vx  = 

—  ka  sin  kt, 

v. 

=  0, 

Vy    = 

Therefore 

ka  cos  kt, 

H 

=  ka. 

fx=  —  k2a  cos  kt 

=  —  k  '-'.r , 

/.-£-"*--»* 

fr  =  V  =  0, 

fy  =  —  k2a  sin  kt 

=  -  Vy, 

*■}>■* 

fn=-'-   = 
P 

_  W 
a 

f  =-k% 

„.-i»-a 

=  -  ka*. 

It  will  be  observed  that  fT  has  a  value  different  from  zero,  while  vT 
is  nil. 

PROBLEMS. 

1.  A  flywheel  making  250  revolutions  per  minute  is  brought  to  rest  in 
2  minutes.      Find  the  average  angular  acceleration. 

2.  A  flywheel  making  250  revolutions  per  minute  is  retarded  by  a 

constanl  acceleration  of  —5-^ — j.     How  many  revolutions  will  the  fly- 
sec.2  J  J 

wheel  make  before  stopping? 

3.  In  the  preceding  problem  find  the  time  it  takes  the  flywheel  to 
come  to  rest. 

4.  Find  the  angular  velocity  and  angular  acceleration  of  a  particle 
which  moves  in  a  manner  defined  by  the  following  pairs  of  equations: 

(a)  p  =  a  sin  cot,  6  =  b  sin  ut. 

(b)  p  =  a  sin  cot,  6  =  b  cos  cot. 

(c)  p  =  a  sin  cot,  d  =  bt. 

(d)  p  =,acbt,  Q  =  bt. 

6.    In  problem  4  find  tha  equation  of  the  path  and  plot  it. 


motion  99 


GENERAL    PROBLEMS. 

Find  the  velocity,  the  acceleration,  and  the  path  of  a  particle  whose 

motion  is  denned  by  the  following  pair  of  equations: 


1. 

X     =  (Hkt, 

y  =  bc-kt. 

2. 

x   =  a  sin  cot, 

y  =  a  sin  2  cot. 

3. 

x   =  a  sin  cot, 

y  =  b  cos  2  u>£. 

4. 

x   =  asm  (cot  +  8), 

?/  =  6  sin  (cot  +  5). 

5. 

x   =  asm  (cot  +  8), 

y  =  bcos  (cot  +  5). 

6. 

x2  =  4  y, 

2/  =  at2. 

7. 

x   =  A  at, 

y  =  bx. 

8. 

x2  =  4  ay, 

y  =  a  sin  co<. 

9. 

x   =  a  cos  cot, 

?/'2  =  4  ax. 

LO. 

x2  +  y2  =  a2,     x2  + 

y2  =  b\ 

CHAPTER   VI. 
MOTION   OF  A   PARTICLE. 

KINETIC    REACTION. 

92.  Kinetic  Reaction. — The  Law  of  Action  and  Reaction, 
which  we  found  so  useful  in  studying  equilibrium,  is  appli- 
cable not  only  to  problems  of  equilibrium  but  also  to  those 
of  motion.  In  applying  it  to  motion,  however,  we  must 
extend  the  meaning  of  the  term  "reaction"  so  as  to  include 
a  form  of  reaction  which  is  known  as  kinetic  reaction.  In 
order  to  understand  the  nature  of  kinetic  reaction  consider 
the  following  ideal  experiment: 

Suppose  you  hold  one  end  of  a  long  elastic  string,  the 
other  end  of  which  is  attached  to  a  rectangular  block  placed 
upon  a  perfectly  horizontal  and  smooth  table.  Let  another 
person  pull  the  block  along  the  plane  of  the  table  and  thereby 
stretch  the  string.  While  the  string  is  being  stretched  you 
have  to  exert  a  force  on  it  in  order  to  keep  your  end  of  it  fixed. 
At  any  instant  the  force  with  which  you  pull  the  string  equals 
and  is  opposite  to  the  force  with  which  the  string  pulls  your 
hand.  The  action  equals  the  reaction  and  is  oppositely  di- 
rected. The  same  is  true  about  the  block  and  the  person  who 
holds  it.  What  will  happen  if  the  block  is  released?  Will 
the  force  which  the  string  exerts  on  your  hand  cease  as  soon 
as  the  block  is  released?  No.  The  string  pulls  on  until  it 
regains  its  natural  length,  something  which  does  not  take 
place  instantaneously.  The  elasticity  of  t  he  string  urges  it  to 
assume  its  natural  Length.  But  this  cannot  be  accomplished 
without  moving  the  block.  Therefore  the  string  moves  the 
block.  Hut  in  order  to  start  the  block  the  string  must  exert 
a  tone  on  it,  and  this  in  spite  of  the  fact  that  the  weight  of 

100 


MOTION  OF  A   PARTICLE  101 

the  block  is  exactly  balanced  by  the  reaction  of  the  plane  so 
that  there  are  no  forces  to  be  overcome  in  order  to  move  tin- 
block.  Therefore  we  conclude  that  the  block  resists  any 
attempt  to  start  it  into  motion.  In  other  words,  the  block 
offers  resistance  to  a  force  which  accelerates  it.  This  resist- 
ance is  the  kinetic  reaction. 

In  one  respect  kinetic  reaction  is  similar  to  frictional  and 
resisting  forces,  namely,  it  is  not  aggressive.  The  kinetic 
reaction  of  a  body  manifests  itself  only  when  its  state  of 
rest  or  of  motion  is  interfered  with.  A  body  which  is  at 
rest  or  moving  with  a  constant  velocity  does  not  display  any 
kinetic  reaction,  but  as  soon  as  it  is  set  in  motion  or  its 
velocity  is  changed  kinetic  reaction  appears;  further  the 
kinetic  reaction  of  a  body  is  greater  the  greater  the  accelera- 
tion imparted  to  it. 

93.  Generalization  of  the  Law  of  Action  and  Reaction.  — 
When  the  terms  "action"  and  " reaction"  are  used  so  as  to 
mean  kinetic  reactions  as  well  as  forces  and  torques,  then  the 
law  is  directly  applicable  to  problems  of  motion  as  well  as 
to  problems  of  equilibrium.  It  will  be  remembered  that  in 
Chapter  III  the  law  was  split  into  two  sections,  of  which  the 
second  section  is  not  applicable  to  single  particles.  There- 
fore we  need  to  consider  here  only  the  first  section,  which 
states : 

To  every  linear  action  there  is  always  an  equal  and  op- 
posite linear  reaction,  or  the  sum  of  all  the  linear  actions 
to  which  a  body  or  a  part  of  a  body  is  subject  at  any  in- 
stant vanishes. 

2A<  =  0.  \ 

If  we  replace  the  term  "linear  action"  by  the  terms 
"force"  and  "linear*  kinetic  reaction"  the  law  may  be  put 
into  the  following  form. 

*  The  adjective  "linear"  is  introduced  in  order  to  distinguish  b< 
the  kinetic  reaction  which  is  celated  to  forces  and  the  kinetic  reaction  intro- 
duced ut  the  beginning  of  Chapter  IX.  which  is  related  to  toi 


102 


ANALYTICAL  MECHANICS 


The  sum  of  all  the  forces  acting  upon  a  body  plus  the 
linear  kinetic  reaction  equals  zero,  or  the  resultant  of  all 
the  forces  acting  upon  a  body  equals  and  is  opposite  to 
the  linear  kinetic  reaction. 

Sum  of  all  forces  +  linear  kinetic  reaction  =  0' 
or    Resultant  force  =  —(linear  kinetic  reaction). 


(A/) 


□ 


□ 


Now  let  us  apply  the  law  to  the  experiment  of  the  preced- 
ing section.  .After  the  block  is  released  it  is  acted  upon  by 
three  forces,  namely,  its  weight,  the  reaction  of  the  table,  and 
the  pull  of  the  string.  The  law  states 
that  the  resultant  of  these  forces  equals 
the  kinetic  reaction  of  the  block  and  is 
oppositely  directed.  Since  the  weight 
and  the  reaction  of  the  table  are  exactly 
balanced  the  pull  of  the  string  is  the 
resultant  force.  Therefore  the  kinetic 
reaction  of  the  block  equals  the  pull  of 
the  string  and  has  a  direction  opposite 
to  that  in  which  the  block  is  pulled. 

94.  Definition  of  Mass.  —  In  the  block 
experiment  suppose  the  free  end  of  the 
string  to  be  connected  to  a  spring  bal- 
ance which  is  fixed  on  the  table,  Fig.  59. 
Further  suppose  the  block  to  be  set  in 
motion  as  in  the  previous  experiment. 
Let  one  person  observe  the  readings  of 
the  balance  and  another  the  acceleration 
of  the  block.  If  Fu  F2,  F3,  etc.,  denote 
the  readings  of  the  balance  and  fh  f2,  /3, 
etc.,  the  wiliics  of  the  acceleration  of  the 
Mock,  obtained  simultaneously  with  the 
readings  of  the  balance,  then  it  will  be  found  that  the  fol- 
lowing relations  hold  true: 

Fi     F2     F8 


Fia.  59. 


m, 


(i) 


MOTION  OF  A   PARTICLE  103 

where  m  is  a  constant.  Bui  since  the  readings  oi  the  balance 
give  the  values  of  the  kinetic  reaction,  equation   1 1  states  that 

the  kinetic  reaction  of  the  block  is  proportional  to  the  acceler- 
ation. The  constant  of  proportionality,  m,  is  called  the  mass 
of  the  block.  We  have,  therefore,  the  following  definition 
for  mass. 

The  mass  of  a  body  is  a  constant  scalar  magnitude  which 
equals  the  quotient  of  the  magnitude  of  the  kinetic  reaction  of 
the  body  by  the  magnitude  of  its  acceleration. 

95.  Measure  of  Kinetic  Reaction.  —Suppose  we  have  sev- 
eral sets  of  apparatus  consisting  of  a  spring;  balance,  a  long 
elastic  string,  and  a  block,  set  up  on  a  smooth  horizontal 
table.  Let  two  persons  attend  to  each  set  of  apparatus:  "in- 
to observe  the  readings  of  the  balance  and  the  other  to 
observe  the  acceleration  of  the  block.  Suppose  the  block- 
to  be  set  in  motion  as  in  the  last  experiment,  and  the  pull 
registered  by  each  balance  observed  at  an  instant  when  the 
corresponding  block  attains  a  certain  definite  acceleration  /. 
Then  if  Fh  F-2,  Fz,  etc.,  denote  the  readings  of  the  balances 
and  mu  m2,  ??i3,  etc.,  the  masses  of  the  blocks,  it  will  be  found 
that  the  following  relations  hold  good: 

E±  =  F±  =  l±=  .  .  .  =/.  ,| 

mx      m%      "h 

Equations  (II)  state  that  when  bodies  have  equal  accelera- 
tions their  kinetic  reaction-  are  proportional  to  their  masses. 
Therefore  equations  (I)  and  (II)  state  that  the  kinetic  re- 
action of  a  body  is  proportional  to  the  product  of  it-  ma—  by 
its  acceleration;  that  is, 

kinetic  reaction  =  kmf,  II I 

where  k  is  the  constant  of  proportionality.  When  the 
quantities  involved  in  the  lasl  equation  are  measured  in  the 
same  system  of  units  the  constant  k  becomes  unity,  in  which 
case  we  have 

kinetic  reaction     mf.  .Ill 


104 


ANALYTICAL  MECHANICS 


If  we  want  to  express  the  fact  that  kinetic  reaction  and  ac- 
celeration are  oppositely  directed  we  put  the  last  equation 
in  the  vector  notation  and  write, 


kinetic  reaction 


mf. 


(IV) 


Equations  (I)  and  (II)  and  consequently  equation  (IV) 
hold  good  not  only  when  the  acceleration  is  due  to  a  change 
in  the  magnitude  of  the  velocity  but  also  when  it  is  due  to 
a  change  in  its  direction.  As  an  illustration  of  this  fact 
consider  the  following  ideal  experiment : 

Let  P,  Fig.  60,  be  a  particle  attached  to  the  end  of  an 
inextensible  string,  which  passes  through  the  hole  0,  in  the 
middle  of  the  smooth  and  hori- 
zontal table  A,  and  is  fastened 
to  the  spring  balance  S.  If  we 
project  the  particle  in  the  plane 
of  the  table  in  a  direction  at 
right  angles  to  the  line  OP  we 
will  find  that  it  describes  a  circle 
about  the  point  0,  with  a  speed 
equal  to  the  speed  of  projection. 
We  will  further  observe  that 
the  balance  registers  a  pull. 

Now  let  us  examine  the  forces  experienced  by  the  particle 
during  its  motion.  The  particle  is  acted  upon  by  three 
forces,  namely,  its  weight,  the  reaction  of  the  table,  and  the 
pull  of  the  string.  Since  the  surface  of  the  table  is  per- 
fectly smooth  and  horizontal  the  weight  and  the  reaction 
of  the  plane  exactly  balance  each  other.  Therefore  the 
pull  of  the  string  is  the  resultant  force.  Thus  the  particle 
ie  pulled  tow  aid  the  point  0,  but  somehow  manages  to  keep 
the  same  distance  from  it;  and  this  in  spite  of  the  fact  that 
it  is  not  acted  upon  by  forces  which  would  counterbalance 
the  pull  of  the  string.  The  explanation  is  plain.  While 
describing  the  circle  the  direction  of  the  velocity  of  the 


/-    4--"  / 

r ! (/ 

i 

ra       ijii r-' 

J 

Fig.  60. 


MOTION  OF  A  PARTICLE  L05 

particle  is  continually  changing,  that  is,  the  particle  is  being 
accelerated.  Therefore  kinetic  reaction  manifests  itself  and 
acts  in  a  direction  opposed  to  that  of  the  acceleration,  thai 
is,  away  from  the  point  0.  Hence  the  pull  of  the  string. 
But  the  pull  which  conies  into  play  is  just  enough  to  over- 
come the  kinetic  reaction,  therefore  the  particle  neither 
approaches  to,  nor  recedes  from,  the  point  0. 

Suppose  we  project  the  particle  with  different  velocities, 
observe  the  corresponding  readings  of  the  spring  balance, 

and  compute  the  accelerations  from  —  —  ,  the  expression  for 

p 
the  normal  acceleration,  p.  94.  Let  F\y  F2,  F3,  etc.,  denote 
the  readings  of  the  balance  and  fl}  /2,  /3,  etc.,  denote  the 
accelerations;  then  we  shall  find  that  the  relations  between 
the  accelerations  and  the  readings  of  the  balance  are  given 
by  equations  (I). 

On  the  other  hand  if  we  fasten  particles  of  different  masses 
to  the  string  and  give  them  equal  accelerations,  we  shall 
find  that  equations  (II)  hold  true.  Therefore  we  conclude 
that  whether  the  acceleration  be  due  to  changes  in  the  mag- 
nitude of  the  velocity,  or  in  the  direction,  or  in  both,  the 
kinetic  reaction  equals  the  product  of  the  mass  by  the 
acceleration  and  is  opposed  to  the  latter. 

The  kinetic  reaction  of  the  last  experiment  may  be  differ- 
entiated from  that  of  the  experiments  of  sections  92  and  '.M 
by  emphasizing  the  fact  that  the  former  comes  into  play 
when  there  is  a  normal  acceleration,  while  the  latter  mani- 
fests itself  whenever  there  is  acceleration  along  the  tangent. 
The  resultant  or  total  kinetic  reaction  is  the  vector  Bum  of 
the  two.  . 

The  results  of  the  last  few  sections  may  be  summed  up  in 
the  following  manner: 

(a)    The  tangential  kinetic  reaction  has  a  magnitude  m'vand 
has  a  direction  opposite  to  that  of  On  tangential  oca  U  ration. 
Tangential  kinetic  reaction  -  -  TOVr.  I\ 


106  ANALYTICAL  MECHANICS 

TO)|2 

(b)  The  normal  kinetic  reaction  has  a  magnitude  and 

p 

has  a  direction  opposite  to  that  of  the  normal  acceleration.  In 
other  words  it  has  the  same  direction  as  the  radius  of  curvature, 
i.e.,  away  from  the  center  of  curvature, 

Normal  kinetic  reaction  = •  (IV") 

P 

/  v4 

(c)  The  total  kinetic  reaction  has  a  magnitude  m  V  v2-\ 

T  p~ 

and  has  a  direction  opposite  to  that  of  the  total  acceleration, 

Total  kinetic  reaction  =  —mv.  (IV) 

FORCE    EQUATION. 

96.   Force  Equation.  —  Combining  (A/)  and  (IV)  and  denot- 
ing the  resultant  force  by  F  we  obtain 

F  =  mi ,  1 
=  mv.l 


(V) 


Equation  (V)  is  called  the  force  equation.  It  states  that  the 
resultant  force  acting  upon  a  particle  equals  the  product  of 
the  mass  by  the  acceleration  and  has  the  same  direction  as 

the  latter. 


/         r4 
IS  V  v2  +  — 

P" 
takes  the  following  form  when  stripped  of  its  vector  notation: 


Since  the  magnitude  of  v  is  V  i>2  +  — ,  the  force  equation 


F  =  m\/i>*-  +  -0.  (VI) 

1  p" 

In  equation  (VI)  v  represents  that  part  of  the  acceleration 
which  is  due  to  the  change  in  the  magnitude  of  the  velocity 

and      represents  that  part  which  is  due  to  the  change  in  the 

p 
direction.* 

97.    Component-force   Equations.  —  Splitting  equation   (V) 
into  two  component  equations  which  correspond  to  the  di- 

p. '.»!  (Or  the  tangential  and  normal  components  of  v. 


MOTION  OF  A  PARTICLE  107 

rections  of  the  tangent  and  the  normal  and  then  dropping 
the  vector  notation  we  obtain 

FT  =  mv,  (VII) 

and  Fn=-m— .  (VIII) 

p 

The  negative  sign  in  equation  (VIII.)  states  that  the  normal 
component  of  the  resultant  force,  and  consequently  the  re- 
sultant force  itself,  is  directed  toward  t he  concave  part  of  t he 
path.  The  last  two  equations  may  be  obtained  directly  from 
(VI)  by  considering  them  as  the  force  equations  for  special 
cases  of  motion.  Thus  when  the  path  of  the  moving  pari  icle 
is  a  straight  line  p  =  oo,  and  consequently 

F  =  mv.  \II\. 

On  the  other  hand  when  the  particle  moves  with  a  con-taut 
speed  v  =  0,  and  therefore 

F  =  -m--  VIII') 

P 

If  in  addition  the  radius  of  curvature  of  the  path  doe-  nol 
change,  that  is,  if  the  particle  moves  in  a  circle  with  a  con- 
stant speed,  then 

F  =  -m->  \  111") 

where  r  is  the  radius  of  the  circle. 

The  following  is  a  useful  set  of  component-force  equations 
obtained  by  splitting  equation  V)  into  three  component 
equations  which  correspond  to  the  directions  of  the  axes  of 

a  rectangular  system: 

Fx  =  nix,  I 

Fv  =  my,\  IX) 

I<\  =  ///:.  I 

Equations  (IX)  emphasize  the  fact  that  ih>  component  of 
the  resultant  force  along  any  direction  equals  (!■•   product  <>f 


108  ANALYTICAL  MECHANICS 

the  mass  by  the  component  of  the  acceleration  along  the  same 
direction. 

98.  Equilibrium  as  a  Special  Case  of  Motion.  — -  When  the 
right-hand  member  of  the  force  equation  vanishes,  that  is, 
when  the  acceleration  is  nil,  the  resultant  force  vanishes. 
But  this  is  the  condition  of  the  equilibrium  of  a  particle, 
therefore  equilibrium  is  a  case  of  motion  in  which  accelera- 
tion is  zero.  For  the  equilibrium  of  a  particle  it  is  neces- 
sary that  the  resultant  force  vanish,  but  this  condition  is  not 
sufficient  because  while  the  acceleration  vanishes  when  F  =  0, 
the  velocity  may  have  any  constant  value.  In  other  words 
a  particle  may  be  in  motion  even  when  the  resultant  of  the 
forces  which  act  upon  it  vanishes.  Therefore  in  order  that  a 
particle  stay  at  rest  not  only  must  the  resultant  of  the  forces 
vanish  but  it  must  be  at  rest  at  the  time  of  application  of 
these  forces. 

99.  Dimensions  of  Force.  —  In  discussing  the  equilibrium 
of  bodies  we  only  compared  forces  because  it  was  all  that  was 
necessary;  besides  we  had  no  means  of  expressing  forces  in 
terms  of  other  physical  magnitudes.  But  now  the  force 
equation  enables  us  to  express  forces  in  terms  of  the  three 
fundamental  magnitudes  and  thus  to  connect  them  with 
other  physical  quantities. 

If  we  substitute  the  dimensions  of  mass  and  acceleration 
in  the  force  equation  we  obtain  the  following  dimensional 
formula  for  force : 

[F]=[MLT~*]. 

100.  Units  of  Force.  —  The  C.G.S.  unit  of  force  is  the  dyne. 

It  is  a  force  which  gives  a  body  of  one  gram  mass  a  unit 

acceleration.     This  is  denoted  symbolically  by  the  following 

formula: 

dyne=,SEi«S:. 

sec.2 

The  British  unit  of  force  is  the  pound,  which  we  have 
already  defined  (p.  76)  as  the  weight,  in  London,  of  a  body 


Motion   ol    a   PARTICLE  109 

which  has  a  mass  of  about  453.G  gins.  The  weight  of  a  body 
is  the  force  with  which  it  is  attracted  toward-  the  center  of 
the  earth.    Therefore  if  m  denotes  the  mass  of  a  body  and 

g  the  magnitude  of  the  acceleration  which  the  gravitational 
attraction  of  the  earth  imparts  to  bodies,  the  force  equation 
gives  us 

W  =  mg,  \ 

where  IF  is  the  weight  of  the  body.  The  value  of  g  is 
slightly  different  at  different  points  of  the  surface  of  the 
earth.  It  is  greatest  at  the  poles  and  least  at  the  equator. 
The  maximum  variation,  however,  is  less  than  one  per  cent ; 
therefore  for  most  purposes  it  may  be  considered  as  constant. 

For  engineering  problems  32.2  — '-    or  981 '-  are  close 

sec-  sec- 

enough  approximations  to  the  actual  value  of  g  in  any 

locality. 

The  relation  between  the  pound  and  the  dyne  may  be 

obtained  by  the  help  of  equation  (X).     Thus 

lib.  =  lpd.  X32.2-^ 
sec.2 

=  32.2 1^4 
sec- 

=  4.45xl0seB^ 

sec. 

=  4.45  X  106  dynes, 

where  "lb."  is  the  symbol  for  the  pound  (weighl  I  and  "pd." 
the  symbol  for  the  mass  of  a  body  which  weighs  one  pound. 
In  order  to  emphasize  the  distinction  between  the  two  they 
are  often  called  pound-weight  and  pound-mass. 

101.  Difference  between  Mass  and  Weight. — The  beginner 
often  finds  it  difficult  to  distinguish  between  the  ma—  of  a 
body  and  its  weight.  He  is  apt  to  ask  Buch  a  question  as 
this,  "When  I  buy  a  pound  of  fruit  what   do  I  get,  one 


110  ANALYTICAL  MECHANICS 

pound-mass  or  one  pound-weight?"*  The  difficulty  is  due 
to  the  fact  that  the  common  methods  for  comparing  the 
masses  of  bodies  make  use  of  their  weights. 

There  are  two  general  methods  by  which  masses  may  be 
compared,  both  of  which  are  based  upon  the  force  equation. 
Let  Fi  and  F2  be  the  resultant  forces  acting  upon  two  bodies 
having  masses  mi  and  m^,  and  /i  and  /2  be  the  accelerations 
produced.     Then  the  force  equation  gives 

Fi  =  mi/i, 
F2  =  m2/2, 

.  mi      Fx  /, 

and  —  =  7T-7' 

(1)  If  the  forces  are  of  such  magnitudes  that  the  accel- 
erations are  equal  then  the  masses  are  proportional  to  the 
forces;  for  when/i  =/2,  the  last  equation  becomes 

Vh  =  Ei. 

ra2      F2 

This  gives  us  a  method  of  comparing  masses,  of  which  the 
common  method  of  weighing  is  the  most  important  example. 

*  This  question  may  be  answered  in  the  following  manner.  "The  fruit 
which  you  net  has  a  mass  of  1  pd.  (about  453.6  gra.)  and  which  weighs  1  lb. 
(about  4.4.")  X  106  dynes).  If  the  fruit  could  be  shipped  to  the  moon  during  the 
the  weight  woul  1  diminish  down  to  nothing  and  then  increase  to  about 
one-sixth  of  a  pound.  The  zero  weight  would  be  reached  at  a  point  about 
nine-tenths  of  the  way  over.  Up  to  thai  position  the  weight  would  be  with 
respeel  to  the  earth,  that  is,  the  fruit  would  be  attracted  towards  the  earth; 
bu1  from  there  on  the  weight  would  be  with  respect  to  the  moon.  The  mass 
of  the  fruit,  however,  would  be  the  same  on  the  earth,  during  the  passage,  and 

on  the  moon.      It   would  be  the  same  with  respect    to  the  moon  as  it  is  with 

respeel  to  the  earth.  Mass  is  an  intrinsic  property  of  matter,  therefore  it 
does  not  change.  Weight  is  the  result  of  gravitational  attraction;  conse- 
quently it  depends  upon,  (a)  the  body  which  is  attracted,  (b)  the  bodies 
which  attract  it.  and  (<•)  the  position  of  the  former  relative  to  the  latter.  It 
18  evident  then  fore  that  when  a  body  is  moved  relative  to  the  earth  its  weight 
changes." 


.Motion   OF  A   PARTICLE  111 

If  W\  and  U">  denote  the  weights  of  two  bodies  of  mac 

and  m-i,  then  by  equation  I  X)  we  obtain 

FTi  =  ///,f/, 

and  —  =  — -, 

where  #  is  the  common  acceleration  due  to  gravitational 

attraction. 

(2)  If  the  forces  acting  upon  the  bodies  are  equal  the 
masses  are  inversely  proportional  to  the  accelerations: 

Vh  =k. 

This  gives  us  the  second  method  by  which  masses  may 
be  compared.  The  following  are  more  or  less  practicable 
applications  of  this  method: 

(a)  Let  A  and  B  (Fig.  61)  be  two  bodies  connected  with  a 
long  elastic  string  of  negligible  mass,  placed  on  a  perfectly 


A 


Fig.  61. 

smooth  and  horizontal  table.  Suppose  the  string  to  be 
stretched  by  pulling  A  and  B  away  from  each  other.  It  is 
evident  that  when  the  bodies  are  released  they  will  be  accel- 
erated with  respect  to  the  table  and  that  the  accelerating 
force,  that  is,  the  pull  of  the  string,  will  be  the  same  for  both 
bodies.  Therefore  if /i  and  /•_•  denote  their  accelerations  al 
any  instant  of  their  motion,  the  ratio  of  their  masses  i- 
by  the  relation 

nh     /i 
(b)  Suppose  the  bodies  whose  masses  are  to  be  compared 


112 


ANALYTICAL  MECHANICS 


to  be  fitted  on  a  smooth  horizontal  rod  (Fig.  62)  so  that  they 
are  free  to  slide  along  it.  If  the  rod  is  rotated  about  a  verti- 
cal axis  the  bodies  fly  away  from  the  axis  of  rotation.  If, 
however,  the  bodies  are  connected  by  a  string  of  negligible 
mass  they  occupy  positions 
on  the  two  sides  of  the  axis, 
which  depend  upon  the  ratio 
of  the  masses.  So  far  as  the 
motion  along  the  rod  is  con- 
cerned, each  body  is  equiv- 
alent to  a  particle  of  the 
same  mass  placed  at  the 
center  of  mass  of  the  body.* 

Suppose,  as  it  is  assumed 
in  Fig.  62,  the  horizontal 
rod  to  be  hollow  and  to  have 
smooth  inner  wall;  further 

suppose  the  centers  of  mass  of  the  given  bodies  to  lie  on  the 
axis  of  the  rod.  Then  if  at  the  center  of  mass  of  each  body 
a  particle  of  equal  mass  is  placed  and  the  two  particles  con- 
nected by  means  of  a  massless  string  of  proper  lengths,  the 
positions  of  the  particles  will  remain  at  the  centers  of  mass 
of  the  given  bodies  even  when  the  rod  is  set  rotating  about 
the  vertical  axis. 

Now  let  mi  and  rth  be  the  masses  of  the  particles  and  fx 
and  /2  their  accelerations  due  to  the  rotation  of  the  tube 
about  the  vertical  axis.  Then  since  the  tensile  force  in  the 
string  is  the  same  at  its  two  ends,  the  forces  acting  upon 
the  particles  are  equal.     Therefore  we  have 

F  =  ffli/i  =  W2/2, 


Fig.  62. 


or 


rrh     fi 


*  For  a  proof  of  this  statement  see  p.  242. 


MOTION  OF  A  PARTICLE  113 

But  if  n  and  r2  denote  the  distances  of  the  particles  from  the 
axis  of  rotation,  and  P  the  period  of  revolution,  then 

/,— tf  — i^S    and     /,=  -£=_< 

n  p-  ra  P2 

Therefore  — '  =  -2 

ma     r, 

gives  the  ratio  of  the  masses  of  the  particles  as  well  as  those 
of  the  given  bodies. 

MOTION  OF  A  PARTICLE  UNDER  A  CONSTANT  FORCE. 
102.    Case  I.     Rectilinear  Motion.  —  Suppose  a  particle  of 
mass  m  to  be  acted  upon  by  a  force  F,  which  is  constant  in 
direction  as  well  as  in  magnitude.     Then  the  force  equation 
gives 

«*-/.  a) 

dv      F      ,  ,_,. 

dt      m 

Since  both  m  and  F  are  constant,  /,  the  acceleration,  is  also 
constant.     Integrating  equation  (I')  once  we  obtain 

v  =  ft  +  c, 

where  c  is  a  constant  to  be  determined  by  the  initial  con- 
ditions of  the  motion.  Let  the  initial  velocity  be  denoted 
by  vQ;   then  v=  v0,  when  t=  0,  therefore  c=  v0  and 

V=Vo  +  ft.  (1) 

ds 
Substituting  -r-  for  v  in  equation  (1)  and  integrating, 

dt 

s=  v0t  +hft2+cf. 
Let     s  =  0,  when  t  =  0;  then  c'  =  0.     Therefore 

s  =Vot+$ft*.  (2) 

Eliminating  t  between  equations  1 1)  and  (2)  we  get 

t>»=Pb»+2/8.  (3) 


114  ANALYTICAL  MECHANICS 

103.  Equations  of  Motion.  —  The  force  equation  and  the 
equations  (1),  (2),  and  (3),  which  connect  v,  s,  and  t  are 
called  equations  of  motion.  The  force  equation  will  be 
called  the  differential  equation  of  motion,  while  those  which 
are  obtained  by  integrating  the  force  equation  will  be  called 
the  integral  equations  of  motion. 

104.  Special  Cases:  A.  Motion  when  the  Force  is  Zero. — 
When  the  force  vanishes  the  acceleration  is  zero.  There- 
fore equations  (1)  to  (3)  become 

v=  v0  =  const., 
s  =  v0t. 

Therefore  the  particle  moves  in  a  straight  path  with  un- 
changing velocity. 

105.  B.  Falling  Bodies.— The  force  experienced  by  a  fall- 
ing body  is  its  weight  mg.  Therefore  the  acceleration  of 
the  motion  is  g,  the  gravitational  acceleration  due  to  the 
attraction  of  the  earth.  So  long  as  the  distance  through 
which  the  body  falls  is  very  small  compared  with  the 
radius  of  the  earth,  g  may  be  considered  to  remain  con- 
stant. Therefore  the  motion  of  falling  bodies  may  be 
treated  as  a  special  case  of  rectilinear  motion  under  a  con- 
stant force.  Hence  the  equations  of  motion  of  a  falling 
body  are  obtained  by  replacing  /  by  q  in  equations  (1)  to 
(3).     Making  this  substitution  we  get 

v  =  t/0+  yt, 
s  =  v0t  +  \  gt2, 
v2=  v02+2gs. 

When  a  body  falls  from  rest  the  initial  velocity  is  zero. 
Therefore  we  must  put  v0=0  in  the  lasl  three  equations 
before  using  them  for  bodies  falling  from  rest. 

When  a  body  is  projected  vertically  upward  the  accelera- 
tion is  in  the  opposite  direction  from  the  velocity:  in  other 
words,  it  is  negative.     Therefore  in  the  last  three  equations 


MOTION  OF  A   PARTICLE  1  i:> 

g  must  be  replaced  by  {-g)  before  they  are  applied  to  the 
motion  of  bodies  which  are  projected  vertically  upwards. 

PROBLEMS. 

1.  A  steel  plate  weighing  10  pounds  is  placed  on  a  perfectly  smooth 
and  perfectly  horizontal  sheet  of  ice.  The  plate  is  then  moved  by  means 
of  a  string,  one  end  of  which  is  fastened  to  the  plate,  and  the  string  is 
pulled  in  a  direction  parallel  to  the  surface  of  the  ice.  Is  it  necessary  to 
apply  a  force  to  the  string  in  order  to  give  the  plate  a  desired  velocity? 
Why?     What  will  the  magnitude  of  the  force  depend  upon? 

2.  In  the  preceding  problem  the  block  is  given  a  velocity  of  LOO  — 

in  5  seconds.     Find  the  tension  of  the  string  supposing  it  to  be  constant . 
How  far  will  the  plate  have  traveled  in  the  meantime? 

3.  In  the  preceding  problem  the  string  is  just  strong  enough  to  support 
half  the  weight  of  the  plate.  What  is  the  shortesl  time  in  which  the 
plate  can  be  pulled  through  a  distance  of  162  feet'.' 

4.  In  the  preceding  problem  suppose  the  contact  to  be  rough  and  to 
have  a  coefficient  of  friction  equal  to  0.1.  A 

5.  A  bullet  is  fired  with  a  muzzle  velocit  yjff 500  metcrs .     Fin- 1  the 

average  acceleration,  supposing  the  length  <j^B  barrel  to  he  si)  cm. 

6.  A  stone  is  sent  gliding  over  a  horizo^^r  sheet  of  ice  with  a 

motors 

of  10 .     How  far  and  how  long  will  it  move  if  the  coefficient  of 

sec. 

friction  is  0.1? 

7.  An  elevator  starts  from  rest  and  rises  to  a  height  of  100  feet  in 
10  seconds,  with  a  constantly  increasing  velocity.  Find  the  increase  in 
pressure  exerted  on  the  feet  of  a  man  in  the  elevator  who  weighs  1"><> 
pounds. 

8.  A  man  can  just  lift  350  pounds  when  on  the  ground.     Bow  much 

can  he  lift  when  in  an  elevator  descending  with  an  acceleration  of  I 

9.  An  elevator  starts  from  rest  and  rises  100  feel  in  five  seconds,  with 
a  constant  acceleration.  Find  the  tension  of  the  rope  which  pulls  it  up 
if  the  elevator  weighs  2000  pounds;   neglect  the  frictional  t"< 

10.  A  body  is  projected  vertically  upward  with  a  velocity  of  50 

the  edge  of  a  pit  200  feet  deep.     When  will  it  strike  the  bottom? 

11.  What  is  the  lowest  level,  over  the  enemy's  camp,  to  which  a  bal- 
loon can  safely  descend,  if  the  enemy  is  provided  with  guns  which  have 
muzzle  velocities  of  2000  feel  per  second? 


116 


ANALYTICAL  MECHANICS 


12.  A  tody,  which  is  dropped  from  the  top  of  a  tower,  strikes  the 
ground  half  a  second  after  it  passes  by  a  window  8-1  feet  above  the  side- 
walk.    Find  the  height  of  the  tower. 

13.  In  the  preceding  problem  find  the  velocity  with  which  the  body 
strikes  the  ground. 

14.  A  man  weighing  150  pounds  is  obliged  to  leave  his  room  by  way 
of  a  window  50  feet  above  the  sidewalk.  He  has  a  rope  which  is  long 
enough  but  cannot  support  more  than  125  pounds.  What  is  the  least 
velocity  with  which  he  can  reach  the  ground? 

15.  A  ball  is  dropped  in  an  elevator  from  a  point  6  feet  above  the  floor 
of  the  elevator.     How  long  will  it  take  to  strike  the  floor  if  the  elevator  is 

descending  with  a  constant  speed  of  10 — -? 

sec. 


106.  C.  Motion  of  a  Particle  along  a  Smooth  Inclined  Plane. 
—  There  are  two  forces  acting  on  the  particle,  its  weight 
and  the  reaction  of  the  plane. 
The  weight  is  wajand  acts 
downwards.  The  rRtction  of 
the  plane,  N,  is  npttnal  to 
the  plane,  because  t^e"4 plane 
is  smooth.  Therefore  setting 
the  components  of  the  kinetic 
reaction  along  and  at  right 
angles  to  the  plane  equal  to 
the  sum  of  the  corresponding  components  of  the  forces  we 
obtain 

dv 


Fig.  63. 


/// 


dt 


mg  sin  a, 


and 


0  =  N  —  mg  cos  a. 


The  last  equation  states  that  forces  along  the  normal  add 
up  to  zero  and  therefore  do  not  affect  the  motion.  We 
have,  therefore,  to  consider  only  the  first  equation,  which 
gives 

dv 

dt 


g  sin  a  =  const. 


MOTION  OF  A  PARTICLE 


117 


Therefore  the  equations  of  motion  are  obtained   by  sub- 
stituting g  sin  a  for/  in  equations  (1)  to  (3).     Thus  we  have 

v  =  v0  +  gt  sin  a, 
s  =  v0t  -\-\gt'1  sin  a, 
v-  =  v0-  +2gs  sin  a, 

for  the  equations  of  motion. 


PROBLEMS. 

1.  A  number  of  particles  slide  down  smooth  inclined  planes  of  equal 

height.     Show  that  the  time  taken  by  each  particle  to  reach  the  base  is 
proportional  to  the  length  of  the  plane  along  which  it  slides. 

2.  Given  the  base  of  an  inclined  plane,  find  the  height  SO  that  the 
horizontal  component  of  the  velocity  acquired  in  descending  it  may  be 
greatest  possible. 

3.  Two  particles  are  projected  simultaneously,  one  up  and  the  oilier 
down  a  smooth  inclined  plane.  Find  the  velocities  of  projection  if  the 
particles  pass  each  other  at  the  middle  of  the  plane. 

4.  Show  that  the  time  taken  by  a  particle  to  slide  down  any  chord 
which  begins  at  the  highest  point  of  a  vertical  circle  is  constant  and  equals 


6- 


here  a  is  the  radius  of  the  circle. 


5.  A  particle  is  projected  down  an  inclined 
plane  of  length  I  and  height  h.  At  the  same 
time  another  particle  is  let  fall  vertically  from 
the  same  point.  Find  the  velocity  of  the  pro- 
jection of  the  first  particle  if  both  strike  the 
base  at  the  same  time. 

6.  A  ship  stands  at  a  distance  d  from  its 
pier.  Show  that  the  length  of  the  chute  which 
will  make  the  time  of  sliding  down  it  a  mini- 
mum is  d  y/2. 

107.  D.  Motion  of  a  Particle  along  a 
Rough  Inclined  Plane. —  The  only  dif- 
ference between  this  problem  and  the 
last  one  is  that  the  reaction  of  the 
plane  is  not  normal  to  the  surface.  On  accounl  of  friction 
the  reaction  R  has  a  component  along  the  plain'.     1  denoting 


118  ANALYTICAL  MECHANICS 

the  frictional  component  of  R  by  F  and  the  normal  com- 
ponent by  N,  and  equating  the  components  of  the  kinetic 
reaction  along  and  at  right  angles  to  the  plane  to  the  sums 
of  the  corresponding  components  of  the  forces  we  obtain 

m  —  =  mg  sin  a  —  F,  (a) 

at 

Q  =  N  —  mg  cos  a.  (b) 

But  if  n  is  the  coefficient  of  friction  then 

F  =  mA7  •  [p.  22] 

=  n  mg  cos  a  [by  (b)]. 

Substituting  this  value  of  F  in  equation  (a)  we  obtain 

dv 
m —  =  mg  sm  a  —  fi  mg  cos  a, 

dv        ,  .  N 

Or  37  =  g  (Sin  a-  ,uCOSa).- 

at 

Thus  the  acceleration  is  constant.  Therefore  the  equations 
of  motion  are  obtained  by  replacing  f  by  g  (sin  a  —  n  cos  a) 
in  equations  (1)  to  (3)  of  page  113: 

v  =  Vq  +  gt  (sin  a  —  n  cos  a), 
s  =  v0t  +  \  gt2  (sin  a  —  fi  COS  a), 
v2  =  vQ2-\-  2  gs  (sin  a  —  n  cos  a). 


PROBLEMS. 

1.  A  car  weighing  10  tons  becomes  uncoupled  from  a  train  which  is 
moving  down  a  grade  of  1  in  200  at  the  rate  of  50  miles  per  hour.  If  the 
frictional  resistance  is  15  pounds  per  ton,  find  the  distance  the  car  will 
travel  before  coming  to  rest. 

2.  The  pull  of  a  locomotive  is  2500  pounds.  Find  the  velocity  it  can 
give  in  5  minutes  to  a  train  which  weighs  75  tons.  Take  10  pounds  per 
ton  for  the  resistance  and  consider  the  (racks  to  be  horizontal. 

3.  In  the  preceding  problem  suppose  the  tracks  to  have  a  grade  of  1 
in  200  and  find  the  velocity  (a)  going  down  grade  and  (b)  going  up  grade. 


MOTION  OF  A  PARTICLE 


119 


108.  E.  Atwood's  Machine.  —  The  problem  is  to  find  tin- 
equations  of  motion  of  two  particles  connected  by  means  ol  a 
string  of  negligible  mass  which  is  slung 
over  a  smooth  pulley. 

Let  mi  and  m2  be  the  masses  of  the 
particles.  Then  considering  each  parti- 
cle separately  we  obtain  the  following 
for  the  force  equations: 

dv 

dt 

dv 

dt 

where  T  is  the  tensile  force  in  the  string. 
Eliminating  T  between  equations  (a) 
and  (b)  we  obtain 

(mi  +  m2)  -£  =  (mi  -  m2)  g,         (c) 


mi 
m2 


T+mtf, 

=  T-m2g, 


dt 

dv  _  mi  —  m2 

dt      mi  +  m2 : 


(d)    t 


ini9 


m2g 


Fig.  65. 


Therefore  the  acceleration  is  constant  and  consequently  tin- 
equations  of  motion  are  obtained  by  substituting  this  value 
of  the  acceleration  in  equations  (1)  to  (3)  of  page  113: 


V 

=  v0  + 

mi-m2 
mi  +  m» 

=  v0t  + 

mi- mi 

& 

2(mi  +  m2)i 

V2 

-*■+ 

mi-  ///•> 
2 ■ gs 

gt\ 


Eliminating  -£  between  equations  (a    and(b)  we  have 


T  = 


2  niit/io 
nh  +  nh 


120  ANALYTICAL  MECHANICS 

Discussion.  —  Instead  of  considering  the  masses  separately  we  can 
consider  them  as  a  single  moving  system  and  write  a  single  force  equation. 
Thus 

(total  moving  mass)  X  (acceleration)  =  sum  of  the  forces, 

or  (mi  +  m2)  -~  =  m^g  -  m^g, 

which  is  identical  with  equation  (c). 

PROBLEMS. 

1.  Two  particles  of  mass  m\  and  ra2  are  suspended  by  a  string  which  is 
slung  over  a  smooth  table.  A  third  particle  of  mass  rn3  is  attached  to  that 
portion  of  the  string  which  is  on  the  table.  Prove  that  when  the  system 
is  left  to  itself  it  will  move  with  an  acceleration  of 


Wi  +  m2  +  ms 

2.  In  the  preceding  problem  suppose  the  table  to  be  rough  and  find 
the  acceleration,     /jl  =  0.5. 

3.  Discuss  Atwood's  machine  supposing  a  frictional  force  to  act  be- 
tween the  string  and  the  pulley  (the  latter  is  supposed  to  be  fixed) ;  take 
the  frictional  force  to  be  equal  to  the  tensile  force  in  that  portion  of  the 
string  which  is  moving  up. 

109.  Case  II.  Parabolic  Motion,  or  Motion  of  Projectiles.  — 
Consider  the  motion  of  a  particle  which  is  projected  in  a 
direction  making  an  angle  a  with  the  horizon.  When  we 
neglect  the  resistance  of  the  air,  the  only  force  which  acts 
upon  the  particle  is  its  weight,  rag.  Taking  the  plane  of 
motion  to  be  the  xy-pl&ne,  Fig.  66,  we  have 

dx      _  dx  Mx 

mit=°>  or  i-°-  (1) 

mt  =  -m9>    or   ft=-g-  (2) 

where    ,X  and   -*  are  the  components  of  the  acceleration 
at  at 

along  the  axes.     Integrating  equations  (1)  and  (2)  we  get 

X—  Ci, 

and  y=  -  gt  +  c2. 


MOTION  OF  A   PARTICLE 


121 


Therefore  the  component  of  the  velocity  along  the  x-axis 
remains  constant,  while  the  component  along  the  {/-axis 
changes  uniformly.  Let  v0  be  the  velocity  of  projection, 
then  when  t  =  0,  x  =  v0  cos  a  and  y  =  v0  sin  a.  Making  these 
substitutions  in  the  last  two  equations  we  obtain 

C\  =  v0  cos  a, 
and 

Therefore  x  =  v0  cos  a,  (3) 

and  ?/  =  v0  sin  a  —  at.  (4) 


(h  =  v0  sin  a. 

X  =  i'o  COS  a, 

y  =  v0  sin  a  - 

-fft. 

A 

"ligt2 

Fig.  66. 

Therefore  the  total  velocity  at  any  instant  is 


v=  Vr^s 


(5) 


(6) 


=  vvQ-  —  2  v0g  sin  a  •  t  +  gH*\ 
and  makes  an  angle  6  with  the  horizon  defined  by 

.      if  Vq  ('OS  a 

tan  0 = *?  =  — -. • 

x       i'0  sm  a  —  #£ 

Integrating  equations  (3)  and  (4)  we  obtain 

X=  r0  COS  a  •/  +  C8j 

and  y  =  v0  sin  a  •  <  —  .*,  g£2  +  c4. 

But   when  £=0,  x  =  y  =  0,    therefore   c3=Q  =  0,  and   con- 
sequently 

x  =  r<) COS  act,  ~ 

y  =  ?'u  sin  act—  \  gt2. 


122  ANALYTICAL  MECHANICS 

It  is  interesting  to  note  that  the  motions  in  the  two  directions 
are  independent.  The  gravitational  acceleration  does  not 
affect  the  constant  velocity  along  the  z-axis,  while  the  mo- 
tion along  the  y-axis  is  the  same  as  if  the  body  were  projected 
vertically  with  a  velocity  v0  sin  a.  The  projectile  virtually 
rises  a  distance  of  v0t  sin  a  on  account  of  its  initial  vertical 
velocity,  and  falls  a  distance  |  gt2  on  account  of  the  gravita- 
tional acceleration. 

The  Path.  —  The  equation  of  the  path  may  be  obtained 
by  eliminating  t  between  equations  (7)  and  (8).     This  gives 

y  =  x  tan  a .  ^    _    x2,  (9) 

which  is  the  equation  of  a  parabola. 

The  Time  of  Flight.  —  When  the  projectile  strikes  the 
ground  its  ^/-coordinate  is  zero.  Therefore  substituting  zero 
for  y  in  equation  (8)  we  get  for  the  time  of  flight 

T=2j^na^  (10) 

g 

The  Range.  — The  range,  or  the  total  horizontal  distance 
covered  by  the  projectile,  is  found  by  replacing  t  in  equation 
(7)  by  the  value  of  T,  or  by  letting  y  =  0  in  equation  (9).  By 
either  method  we  obtain 

2  v02  sin  a  cos  a 


R 


?'02  sin  2  a 
9 


(ID 


Since  v0  and  g  are  constants  the  value  of  R  depends  upon  a. 
It   is  evident  from  equation  (11)  that  R  is  maximum  when 

Bio  2  a=  1,  or  when  a  =  \     The  maximum  range  is,  there- 
fore, 

Rm  =  Vj?.  (12) 

9 


MOTION  OF  A   PARTICLE 


123 


In  actual  practice  the  angle  of  elevation  which  gives  the 
maximum  range  is  smaller  on  account  of  the  resistance  of 
the  air. 

The  Highest  Point.  —  At  the  highest  point  y  =  0.     There- 
fore substituting  this  value  of  ij  in  equation  (4)  we  obtain 

— or  -  T  for  the  time  taken  to  reach  the  highesl  point. 

9  2 

Substituting  this  value  of  the  time  in  equation  (8)  we  gel 
for  the  maximum  elevation 


// 


t'o-sin-q. 
2<7 


(13) 


The  Range  for  a  Sloping  Ground.  —  Let  0  be  the 
angle  which  the  ground  makes  with  the  horizon.  Then  the 
range  is  the  distance  OP,  Fig.  67,  where  P  is  the  point 
where  the  projectile  strikes  the  sloping  ground.  The  equa- 
tion of  the  line  OP  is 

2/ =  :c  tan  0.  (14) 


Eliminating  y  between  equations  (14)  and  (9)  we  obtain  the 

^-coordinate  of  the  point, 

2  /'q2  cos2  a  (tan  a  —  tan  0) 


But 

Therefore 


xp  =  R'  cos  0,     where     W  =  OP. 

^2'^%in(«-0) 
g cos2  0 

_  t'02sin  (2  a  -  0)  -  sin  /3 

g  cos2/3 


(15) 


124  ANALYTICAL  MECHANICS 

Thus  for  a  given  value  of  /3,  R'  is  maximum  when  sin  {2a— 0) 

=  1,  that  is,  when  «  =  t  +  ?: 

4        Z 

,      v02 1  -  sin  ^  v02 

gr     cos-/3         gi(l  +  sin/3) 

When  0  =  0  equations  (15)  and  (16)  reduce  to  equations  (12) 
and  (13),  as  they  should. 


PROBLEMS. 

1.  A  body  is  projected  horizontally  with  a  speed  of  105      -  from  a  cliff 

sec. 

365  feet  high.  Find  the  magnitude  and  direction  of  the  velocity  at  the 
time  it  reaches  the  ground. 

2.  The  muzzle  velocity  of  a  gun  is  3000  — - .     Find  the  area  it  covers 

sec. 

if  it  is  mounted  on  top  of  a  hill  500  feet  above  the  surrounding  plain. 

3.  A  shot  fired  horizontally  from  the  top  of  a  tower  strikes  the  ground 
at  a  distance  d  from  the  base  of  the  tower,  with  a  velocity  the  vertical 
component  of  which  equals  the  initial  velocity  of  the  shot.  Find  the 
height  of  the  tower. 

4.  A  bullet  is  projected  at  an  angular  elevation  of  45°  with  a  velocity 

of  400  — - .     At  the  highest  point  of  its  flight  the  bullet  goes  through  a 

target  5  cm.  thick  and  strikes  the  ground  at  a  distance  of  1200  m.  from 
the  place  where  it  was  projected.  Find  the  average  resisting  force  offered 
by  the  target. 

6.  After  sliding  200  m.  down  a  slope  of  30°  a  ski-jumper  leaves  the 
ground  making  45°  with  the  horizon  and  lands  further  down  the  same 
doping  ground.  Supposing  the  coefficient  of  friction  to  be  0.05  and 
neglecting  the  resistance  due  to  air,  find  (a)  the  speed  with  which  he  left 
the  ground,  (b)  the  speed  with  which  he  landed. 

6.  In  the  preceding  problem  find  the  leap  measured  along  the  ground. 

7.  A  man  can  make  6  feet  in  the  high  jump.  How  many  feet  could  he 
make  if  he  were  on  the  moon?  The  gravitational  acceleration  on  the  moon 

is  5.3  — -  . 
sec.2 

8.  A  particle  slides  down  a  chord  of  a  vertical  circle  and  then  falls  on  a 
horizontal  plane  A  feel  below  the  Lower  end  of  the  chord.  Find  the  chord 
which  will  give  the  greatest  possible  range  on  the  plane. 


MOTION  OF  A  PARTICLE 


125 


9.  Show  that  a  rifle  will  shoot  three  times  as  high  when  it-  angle  of 
elevation  is  60°  as  when  it  is  30°,  but  will  carry  the  Bame  distance  along  a 
horizontal  plane. 

10.  An  emery  wheel  bursts  into  small  pieces  while  making  100 
lutions  per  second.  If  the  radius  of  the  wheel  is  10  cm.  find  farthi 
tance  reached  by  any  of  its  pieces. 


MOTION    OF  A   PARTICLE    UNDER   THE  ACTION   OF   A    VARY 
ING   FORCE. 

110.  I.  Uniform  Circular  Motion.  —  Consider  the  motion  of 
a  particle  projected  into  a  circular  tube,  Fig.  68,  the  inner 
surface  of  which  is  perfectly 
smooth.  Let  m  be  the  mass  of 
the  particle,  v  its  speed  of  pro- 
jection and  r  the  radius  of  the 
circle  formed  by  the  tube.  The 
radius  of  the  cross-section  of 
the  tube  is  supposed  to  be  neg- 
ligible. Suppose  the  particle  to 
be  acted  upon  by  no  forces  ex- 
cept the  reaction  of  the  inner 
surface  of  the  tube.  Then,  since 
the  surface  is  smooth,  the  reac- 
tion is  normal  and  consequently  there  is  no  force  along  the 
tangent  to  the  path.  Therefore  the  force  equations  for  the 
tangential  and  the  normal  directions  give 

do 


Fra.  68. 


and 


"a"0- 


m—  =  —  N, 
r 


(1) 


2 


where  N  represents  the  reaction  of  the  walls  of  the  tube  upon 
the  particle.  It  is  evident  from  equation  (1)  thai  the  mag- 
nitude of  the  velocity  is  constant.  Therefore  the  particle 
describes  the  circular  path  with  a  constant  speed.  Equation 
(2)  states  that  the  normal  reaction  of  the  surface  equals  the 


126  ANALYTICAL   MECHANICS 

normal  kinetic  reaction  and  is  oppositely  directed.  There- 
fore the  reaction  of  the  surface  is  directed  towards  the  center 
of  the  circular  path. 

Equations  (1)  and  (2)  are  independent  of  the  special 
method  used  to  keep  the  particle  in  a  circular  path.  If, 
for  instance,  the  particle  were  connected  to  the  center  by 
means  of  an  inextensible  string  and  then  projected  in  a 
direction  perpendicular  to  the  string  the  results  would  have 
been  the  same. 

The  force  which  constrains  the  particle  to  move  in  the 
circular  path  is  called  the  central  force.  This  force  may  be 
the  reaction  of  a  surface,  the  tension  of  a  string,  or  the 
pull  of  a  center  of  attraction.  In  order  to  emphasize  the 
fact  that  this  force  is  directed  towards  the  center  it  is  often 
denoted  by  Fc.  Since  the  subscript  makes  clear  the  fact 
that  the  force  is  directed  towards  the  center,  we  can  drop 
the  negative  sign  from  equation  (2),  and  write 

ft-2?  (3) 

4?r2mr 
--pi-' 

where  P  is  the  time  of  one  revolution. 


(4) 


PROBLEMS. 

1.  A  particle  of  mass  mi,  which  describes  a  circle  on  a  perfectly  smooth 
horizontal  table,  is  connected  with  another  particle  of  mass  tn2  which 
hangs  freely;  the  string  which  connects  the  two  particles  passes  through  a 
smooth  hole  in  the  table.    Find  the  condition  necessary  to  keep  7n2  at  rest. 

2.  Find  the  smallest  horizontal  velocity  with  which  a  body  must  be 
projected  at  the  equator  in  order  that  the  body  may  become  a  satellite. 
Find  the  period  of  revolution., 

3.  A  locomotive  weighing  125  tons  moves  in  a  curve  of  600  feet  radius, 
with  a  velocity  of  20  miles  per  hour.  Find  the  lateral  pressure  on  the 
rails  if  they  are  on  the  same  level. 

4.  Derive  the  expression  for  the  period  of  a  conical  pendulum. 

6.  A  number  of  particles  of  different  masses  are  suspended  from  the 
same  point  by  means  of  strings  of  different  lengths.     li\ow  that  when 


.MOTION"  OF  A   PARTICLE  127 

the  bodies  are  given  the  same  angular  velocity  about  a  vertical  axis 
through  the  point  of  suspension  the  particles  will  lie  in  the  same  hori- 
zontal plane. 

6.  If  the  masses  in  the  preceding  problem  are  equal  how  will  the  tensile 
force  vary  with  the  length  of  the  strings'.' 

7.  Supposing  the  earth  to  be  spherical  discuss  the  variation  in  the 
weight  of  a  body  due  to  the  rotation  of  the  earth  about  its  axis. 

8.  The  moon  describes  a  circular  path  around  the  earth  once  in  every 
27  days,  7  hours,  and  43  minutes;  find  the  acceleration  at  the  center  of 
the  moon  due  to  the  attraction  of  the  earth.  Take  2-40,000  miles  f,,r  the 
radius  of  the  moon's  orbit. 

9.  If  the  earth  rotated  fast  enough  to  make  the  weights  of  bodies 
vanish  at  the  equator  show  that  the  plumb  line  at  any  latitude  would 
become  parallel  to  the  axis  of  the  earth. 

10.  In  the  preceding  problem  what  would  the  length  of  the  day  be? 

11.  How  much  would  the  weight  of  a  body  be  increased  at  latitude  30° 
if  the  earth  stopped  rotating? 

12.  A  particle  suspended  from  a  fixed  point  by  a  string  of  length  / 
is  projected  horizontally  with  a  speed  Vllg;  show  that  the  string  will 
become  slack  when  the  particle  has  risen  to  a  height  I  I. 

13.  How  much  should  the  outer  rail  of  a  railroad  track  be  raised  at  a 
curve  in  order  that  there  be  no  lateral  pressure  on  the  rails  when  a  train 
makes  the  curve  at  the  rate  of  a  mile  a  minute?  The  radius  of  the  curve 
is  1500  feet  and  the  distance  between  the  tracks  is  4  feet  SJ  inches. 

14.  Prove  that  a  locomotive  will  upset  if  it  takes  a  curve  with  a 

greater  thanl/^,  on  tracks  the  outer  rails  of  which  are  not  raised, 

>   2  h 
where  g  denotes  the  gravitational  acceleration,  r  the  radius  of  the  curve, 
a  the  distance  between  the  rails,  and  h  the  height  of  the  center  of  mass 
of  the  locomotive  above  the  tracks. 

16.    Show  that  if  there  is  no  lateral  pressure  on  the  outer  rails,  while 

a  car  takes  a  curve,  the  relation 

v- 

tanfl  =- 

gr 

is  satisfied,  where  6  is  the  angle  the  floor  of  the  car  makefl  with  the  hori- 
zon, v  is  the  speed  of  the  car,  and  r  the  radius  of  the  curve. 

111.  II.  Bodies  Falling  from  Great  Distances.  -When  the 
distance  from  which  a  body  falls  is  not  negligible  compared 
with  the  radius  of  the  earth  the  gravitational  acceleration 
cannot  be  considered  as  constant  during  the  hill.     Therefore 


128  ANALYTICAL  MECHANICS 

the  variation  of  the  gravitational  attraction  must  be  taken 
into  account.  According  to  the  law  of  gravitational  attrac- 
tion the  force  between  two  gravitating  spherical  bodies  is 

of  the  following  form: 

mm'  m 

f=  -7-^->  (i) 

where  m  and  m'  are  the  masses  of  the  spheres,  r  is  the  dis- 
tance between  their  centers,  and  7  is  a  positive  constant. 
The  negative  sign  indicates  the  fact  that  r  is  measured  in  a 
direction  opposed  to  that  in  which  F  acts.  When  the  grav- 
itating bodies  are  the  earth  and  a  body  which  is  small  com- 
pared with  the  earth  7  =  ^— ,  where  M  is  the  mass  of  the 

earth,  a  its  radius,  and  g  the  gravitational  acceleration  on 
the  surface  of  the  earth.  In  order  to  show  this  observe 
that  when  the  body  is  on  the  surface  of  the  earth,  that 
is,  when  r  =  a,  the  force  is  —  mg,  the  weight  of  the  body. 
Therefore  replacing  in  equation  (1)  F  by- mg  and  m'  by  M 
and  solving  for  7  we  obtain 


ga2 
7      .1/' 

(2) 

Substituting  in  equation  (1)  this  value  of  7  we  get 

f=       mga2 

r- 

(3) 

for  the  force  which  acts  upon  a  body  of  mass  m  during  its 
fall  towards  the  earth. 

Therefore  the  force  equation  is 

dv          mga2                                     ... 

I  dropping  ///  from  both 

sides  of  equation  (4)  and  writing 

dv 
"dr 

for   ,,  we  obtain 
dt 

dv         qa2 
dr          r2 

(5) 

MOTION  OF  A  PARTICLE  129 

Separating  the  variables  and  integrating  we  have 
r-  =  — 1-  c. 

Now  suppose  the  body  starts  to  fall  from  a  distance  r'  from 

the  center,  then  v  =  0  when  r  =  r'  and  c  =  —  *-rm     Therefi  »re 

/• 

--»».»(* -I)  (6) 

gives  the  velocity  at  a  distance  r  from  the  center.  WheD 
the  body  falls  from  an  infinite  distance  r'  =  oc  and  the 
velocity  at  any  distance  is 

,„  =  a\/^.  (7) 

Therefore  the  velocity  with  which  it  will  reach  the  surface 
of  the  earth  is 

Woo  =  ^2  ga 

. .  m  miles    \  (8) 

=  7 

sec.     j 

If  the  body  starts  to  fall  from  a  distance  above  the  surface 
equal  to  the  radius  of  the  earth,  then  in  equation  (6)  r'  =  2r. 
Therefore 

v=  ^ag, 

miles 

=  4.95 

sec. 

Therefore  about  seventy-one  per  cent  of  the  velocity  attained 
in  falling  from  an  infinite  distance  is  developed  in  the  lasl 
4000  miles. 

PROBLEMS. 

1.  A  meteorite  falls  n.  the  earth.     Supposing  it  in  -tart  from  infinity 
find  the  time  it  takes  to  travel  tin-  last    WOO  mile-. 

2.  A  particle  is  attracted  towards  a  fixed  point  by  a  force  which  varies 

inversely  as  the  cube  of  the  distance  of  the  particle  from  the  fixed  point. 
Find  the  time  it  will  take  the  particle  to  fall  to  the  point  if  it  .-tarts  from 
a  distance  d. 


130  ANALYTICAL   MECHANICS 

3.  Discuss  the  motions  of  a  particle  which  is  repelled  from  a  fixed 
point  with  a  force  which  varies  directly  as  the  distance  of  the  particle 
from  the  fixed  point. 

112.  III.  Motion  of  a  Particle  in  a  Resisting  Medium.  —  As 
a  concrete  example  of  motion  in  a  resisting  medium  consider 
the  motion  of  falling  bodies,  taking  the  resistance  of  the 
atmosphere  into  account.  At  any  instant  of  the  motion 
two  forces  act  on  the  body,  i.e.,  the  weight  of  the  body  and 
the  resistance  of  the  air.  Denoting  the  resisting  force  by 
F  we  get 

mjt=mg-F  (1) 

for  the  force  equation.  In  order  to  be  able  to  integrate  the 
last  equation  we  must  make  an  assumption  as  to  the  nature 
of  F. 

Case  I.  Resistance  Proportional  to  the  Velocity. 
—  Suppose  F  to  be  proportional  to  the  velocity,  then 

F=  fav, 

where  ki  is  a  positive  constant.  Substituting  in  the  force 
equation  this  expression  for  F  we  obtain 

dv  , 

"•■■5 -**-**. 

dv  ,  /r,s 

a"'-*  (2) 

h 

where  k  =  — .     Rearranging  the  last  equation 


m 


Integrating 


dv        _  j    if 


Iog(»-|V=-Arf+c, 


or  v  -  |  =  ec  •  e" 


MOTION  OF  A  PARTICLE 


131 


Let    v  =  i'o  when  t  =  0,  then  ec  =  v0  -  j.      Therefore 


or 


+   I 


(3) 


Limiting  Velocity.  —  The  last  equation  has  a  simple  in- 
terpretation which  comes  out  clearly  by  plotting  the  time  as 
abscissa  and  the  velocity  as  ordinate.  There  are  four  special 
cases  which  depend  upon  the  following  values  of  the  initial 
velocity : 


(a)  v0=  0, 


(b)  »0<fi 


(c)    I'o 


I  (d)"»>z- 


Curves  (a),  (b),  (c),  and  (d)  of  Fig.  69  represent  these  ca»  3. 

It  is  evident  from  these  curves 
that  whatever  its  initial  value 
the  velocity  tends  to  the  same 

limiting  value  |,  called  the  lim- 
iting velocity.  In  the  third  case 
the  velocity  remains  constant,  as 
shown  by  the  horizontal  line  (c), 
because  the  resisting  force  ex- 
actly balances  the  moving  force. 
Integrating  equation  (3)  we 
get 


I  i...  60. 


=  h 


rj:  -  g 


k  k 

Let  s  =  0  when  t  =  0,  then 

c  = 


+  c. 


t'ok  -  g 


Therefore 


t  + 


vpk  -  g 
k2 


(l  -  «-*'). 


^ 


132 


ANALYTICAL  MECHANICS 


If  we  plot  the  last  equation  for  the  four  different  cases  of 
the  initial  velocity  we  obtain  the  curves  of  Fig.  70.  It  is 
evident  from  these  curves  that  a  very- 
short  time  after  the  beginning  of  the 
motion  the  distance  covered  increases 
at  a  constant  rate,  as  would  be  ex- 
pected from  the  meaning  of  the  lim- 
iting velocity. 

Case  II.  Resistance  Propor- 
tional to  the  Square  of  the  Ve- 
locity. —  The  assumption  that  resist- 
ance varies  as  the  velocity  holds  only 
for  slowly  moving  bodies.  It  is  found 
that  for  projectiles  whose  velocities 
lie  under  1000  feet  per  second  and 
over  1500  feet  per  second  the  resistance  varies,  approxi- 
mately, as  the  square  of  the  velocity,  while  between  these 
values  it  varies  as  the  cube  and  even  higher  powers  of 
the  velocity.  The  experimental  data  on  the  subject  are 
not  enough  to  find  a  law  of  variation  which  holds  in  all 
cases. 

If  we  assume  the  resistance  to  vary  as  the  square  of  the 
velocity,  then  the  force  equation  for  a  falling  body  becomes 

dv  ,    2 

m  —  =  mg  -  kiv2, 
at 


dv 
dt 


g  -  kv2, 


(5) 


/.-, 


where  k  =  —  =  constant.     In  order  to  integrate  the  last  equa- 

i/i 

t  ion  wo  replace  —  by  v-  and  rearrange  the  terms  so  that  we 
dt  ds 

get 


dv 


kds. 


MOTION  OF  A  PARTICLE  l:;.; 

Integrating  we  have 

\og(v°--gk)=-2ks  +  c, 


or  v-  —  f  =  ec  >e~-k\ 

Let  v  =  vQ  when  s  =  0,  then  ec  =  v02  -  |.    Therefore 


Therefore  the  limiting  velocity  is  y  |.  In  other  words,  for 
large  values  of  s  the  distance  traversed  approximately  equals 

PROBLEMS. 

1.  A  man  finds  that  the  resistance  of  the  air  to  a  body  moving  at  the 
rate  of  20  - — '■  equals  1000  dynes  per  square  centimeter  of  the  resisting  sur- 
face.    If  600 '■  is  the  limit  of  the  velocity  with  which  he  can  safely  land, 

sec. 

find  the  smallest  parachute  with  which  he  ran  safely  descend  from  any 
height.  The  man  and  his  parachute  have  a  mass  of  7.">  kg.  Take  the  re- 
sistance to  be  proportional  to  the  velocity. 

2.  In  the  preceding  problem  take  the  resistance  to  be  proportional  to 
the  square  of  the.  velocity. 

3.  Discuss  the  equation  of  motion  of  a  boal  in  still  water,  after  the. 
man  who  was  rowing  ships  his  oars.  Suppose  the  resistance  to  be  pro- 
portional to  the  velocity. 

4.  A  particle  is  projected  with  a  velocity  V  in  a  rc-i-ting  medium  and 
is  acted  upon  by  no  other  force  except  that  due  to  the  resistance  of  the  m<^ 
dium.  Show  that,  (a)  the  particle  will  describe  a  finite  distance  in  an  in- 
finite time  when  the  resistance  is  proportional  to  the  velocity;  (b)  it  will 
describe  infinite  distance  in  infinite  time  when  the  resistance  i-  propor- 
tional to  the  square  of  the  velocity. 

6.  A  bullet  is  projected  vertically  upwards  with  a  velocity  v,.  Sup- 
posing  the  resistance  of  the  air  to  be  proportional  to  the  square  of  the 


134 


ANALYTICAL    MLCHAXK  \S 


velocity  of  the  bullet,  find  the  expression  for  the  highest  point  reached. 
Also  find  the  time  of  upward  flight. 

6.  In  the  preceding  problem  suppose  the  resistance  to  be  proportional 
to  the  velocity. 

113.  IV.  Simple  Harmonic  Motion. — The  motion  of  a  par- 
ticle is  called  simple  harmonic  when  the  particle  is  acted 
upon  by  a  force  which  is  always  directed  towards  a  fixed 
point  and  the  magnitude  of  which  is  proportional  to  the 
distance  of  the  particle  from  the  same  point. 


Fig.  71. 


Let  m  be  the  mass  of  the  particle,  the  line  A  A'  its  path, 
Fig.  71,  and  0  the  fixed  point.  Then  denoting  the  dis- 
placement, i.e.,  the  distance  of  the  particle  from  the  fixed 
point,  by  x  we  obtain 

F  =  -  k2x, 

dv  ,  „ 

or  m~Ti  =  ~  *  x> 


,11 


(1) 


for  the  force  equation.  It  is  evident  that  equation  (1)  is 
nothing  more  or  less  than  the  analytical  expression  for  the 
foregoing  definition  of  simple  harmonic  motion.  The  fac- 
tor /.-  is  a  constant.  The  negative  sign  in  equation  (1) 
indicates  the  fact  thai  the  force  and  the  acceleration  are 
directed  towards  the  fixed  point,  while  x  is  measured  from 


MOTION  OF  A   PARTICLE  i:;:, 

it.     Since  the  motion  is  along  the  x-axia  the  velocity  has 
no  components  along  the  other  axes,  consequently 
Therefore  equation  (1)  may  be  written  in  the  form 


(2) 


ih.v  , ., 

m——y  =  —  k2x. 

«■  dt- 

where  o>2  =  —  •    Multiplying  both  sides  of  equation  (2)  by 

dx 
%-rrdt  and  integrating 


or  v  =  v  c 


v=Vc 


where   c2  is  the  constant  of  integration.     Let  v  =t^  when 
x  =  0,  then  c~  =  v02.     Therefore 


V=  Vi)02  — co2X2. 

In  order  to  find  the  second  integral  of  equation  (2)  rewrite 
equation  (3)  in  the  form 

^-  =  Vv0*-o>2x\ 
at 

Separating  the  variables  in  the  last  equation  and  integrating 
we  have 

sin  J  —  =  ut  +  c  , 
or  x  =  —  sin  (<at  -f  c') 

CO 

=  a  sin  (ut  +  c')f 

where  c'  is  the  constant   of   integration   and   a  =  -.     Let 

u 

x  =  0  when  £  =  0,  then  c'  ■=  0,  and 

•  x  =  a  sin  cot.  (4) 


136  ANALYTICAL  MECHANICS 

When  equation  (4)  is  plotted  with  the  time  as  abscissa  and 
the  displacement  as  ordinate  the  well-known  sine  curve  is 
obtained,  Fig.  71. 

It  is  evident  both  from  equation  (4)  and  from  the  curve  that 
the  maximum  value  of  x  is  equal  to  a.  This  value  of  the  dis- 
placement is  called  the  amplitude.  The  minimum  value  of 
x  is  a  displacement  equal  to  the  amplitude  in  the  negative 
direction.  Therefore  the  particle  oscillates  between  the 
points  A  and  A'.  The  displacement  equals  the  positive 
value  of  the  amplitude  every  time  sin  ut  equals  unity,  that 

is,  when  at  assumes  the  values  ->  — ,  — ,  etc.     In  other 

2        Z         2 

words,  the  particle  occupies  the  extreme  point  A  at  the 

instants  when  t  has  the  values-^-,  — — ,  — ,  etc.     Therefore 

2u     2  co     2  co 

the  particle  returns  to  the  same  point  after  a  lapse  of  time 
equal  to  — — .     This  interval  of  time  is  called  the  period  of 

CO 

the  motion  and  is  denoted  by  P.     Thus 

P-—  (5) 


PROBLEMS. 

1.  A  particle  which  moves  in  a  straight  groove  is  acted  upon  by  a  force 
which  is  directed  towards  a  fixed  point  outside  the  groove,  and  which 
varies  as  the  distance  of  the  particle  from  the  fixed  point.  Show  that  the 
motion  is  harmonic. 

2.  Within  the  earth  the  gravitational  attraction  varies  as  the  dis- 
tance from  the  center.  Find  the  greatest  value  of  the  velocity  which  a 
body  would  attain  in  falling  into  a  hole,  the  bottom  of  which  is  at  the 
center  of  the  earth. 

3.  Show  that  when  a  particle  describes  a  uniform  circular  motion,  its 
projection  upon  a  diameter  describes  a  harmonic  motion. 

GENERAL  PROBLEMS. 

1.  The  speed  of  a  train  which  moves  with  constant  acceleration  is 
doubled  ID  a  distance  of  3  kilometers.  It  travels  the  next  1  iV  kilometers 
in  one  minute.     Find  the  acceleration  and  the  initial  velocity. 


MOTION  OF  A  PARTICLE  137 

2.  Show  that  the  ratio  of  the  distances  described  by  a  falling  body 

during  the  (n  —  l)th  and  the  nth  seconds  is  ~  "— 1  . 

2«  +  l 

3.  A  juggler  keeps  three  balls  going  in  one  hand,  bo  thai  al  any  in- 
stant two  arc  in  the  air  and  one  in  his  hand.  Find  the  time  during  which 
a  ball  stays  in  bis  hand;  each  ball  rises  to  a  height  h. 

4.  Find  the  shortest  time  in  which  a  mass  m  can  be  raised  to  a  height  h 
by  means  of  a  rope  which  can  bear  a  tension  T. 

6.  A  train  passes  another  on  a  parallel  track.  When  the  two  loco- 
motives are  abreast  one  of  the  trains  has  a  velocity  of  20  miles  per  hour 

and  an  acceleration  of  3  — -OJ  while  the  other  has  a  velocity  of  40  miles 
sec- 

pcr  hour  and  an  acceleration  of  1  — '—.    How  soon  will  they  be  abreast 

sec.2 

again,  and  how  far  will  they  have  gone  in  the  meantime0 

6.  A  mass  of  1  kg.  is  hanging  from  a  spring  balance  in  an  elevator. 
After  the  elevator  starts  the  balance  reads  1100  gin.  Assuming  the 
acceleration  of  the  elevator  to  be  constant,  find  the  distance  moved  in 
5  seconds. 

7.  A  smooth  inclined  plane  of  mass  m  and  inclination  a  stands  with 
its  base  on  a  smooth  horizontal  plane.  What  horizontal  force  must  be 
applied  to  the  plane  in  order  that  a  particle  placed  mi  the  plane  simulta- 
neously with  the  beginning  of  action  of  the  force  may  lie  in  contact  with 
the  plane  yet  fall  vertically  down  as  if  the  inclined  plane  were  not  then'? 

8.  The  pull  of  a  train  exceeds  the  resisting  forces  by  0.02  of  tin'  weight 
of  the  train.  When  the  brakes  are  on  full  the  resisting  forces  equal  0.1 
of  the  weight.  Find  the  least  time  in  which  the  train  can  travel  between 
two  stopping  stations  5  miles  apart,  the  tracks  being  leveL 

9.  Give  a  construction  for  finding  the  line  of  quickest  descent  from  a 
point  to  a  circle  in  the  same  vertical  plane. 

10.  A  mass  mx  falling  vertically  draws  a  mass  mt  up  a  smooth  inclined 
plane  which  makes  an  angle  of  30°  with  the  horizon.     The  un- 
connected with  a  string  which  passes  over  a  -mall  smooth  pulley  at  the 
top  of  the  plane.     Find  the  ratio  of  the  masses  which  will  make  the 

acceleration  *■  • 
4 

11.  A  particle  is  projected  up  an  inclined  plane  which  makes  an  angle 
a  with  the  horizon.     If  7',  is  the  time  of  ascent,  T .  the  time  of  d 

and  </>  the  angle  of  friction,  show  that 


(TX-  _  sin  (a  -  4>) 


'   r-  <t>) 


138  ANALYTICAL  MECHANICS 

12.  The  time  of  descent  along  straight  lines  from  a  point  on  a  verti- 
cal circle  to  the  center  and  to  the  lowest  point  is  the  same.  Find  the 
position  of  the  point. 

13.  A  uniform  cord  of  mass  m  and  length  I  passes  over  a  smooth  peg 
and  hangs  vertically.     If  it  slides  freely,  show  that  the  tension  of  the 

cord  equals     'r    /2~      ,  when  the  length  on  one  side  is  x. 

14.  In  an  Atwood's  machine  experiment  the  sum  of  the  two  moving 
masses  is  m.  Find  their  values  if  in  t  seconds  they  move  through  a  dis- 
tance h. 

15.  Given  the  height  /;  of  an  inclined  plane,  show  that  its  length  must 

be  — -^r,  in  order  that  mi,  descending  vertically,  shall  draw  m-i  up  the  plane 
mji 

in  the  least  possible  time. 

16.  A  gun  points  at  a  target  suspended  from  a  balloon.  Show  that 
if  the  target  be  dropped  at  the  instant  the  gun  is  discharged,  the  bullet 
will  hit  the  target  if  the  latter  is  within  the  bullet's  range. 

17.  Find  the  position  where  a  particle  sliding  along  the  outside  of  a 
smooth  vertical  circle  will  leave  the  circle. 

18.  A  particle  falls  towards  a  fixed  point  under  the  action  of  a  force 
which  equals  yr~ri,  where  7  is  a  constant  and  r  is.  the  distance  of  the  parti- 
cle from  the  fixed  point.    Show  that  starting  from  a  distance  a  the  particle 

will  arrive  at  the  fixed  point  with  an  infinite  velocity  in  the  time       • 

V3t 

19.  A  particle  falls  towards  a  fixed  point  under  the  attraction  of  a  force 
which  varies  with  some  power  of  the  distance  of  the  particle  from  the  cen- 
ter of  attraction.  Find  the  law  of  force,  supposing  the  velocity  acquired 
by  the  particle  in  falling  from  an  infinite  distance  to  a  distance  a  from  the 
center  to  be  equal  to  the  velocity  acquired  in  falling  from  rest  from  a  dis- 
tance a  to  a  distance  -• 

4 

20.  A  part  icle  is  projected  toward  a  center  of  attraction  with  a  velocity 
equal  to  the  velocity  it  would  have  acquired  had  it  fallen  from  an  infinite 
distance  to  the  position  of  projection.  Supposing  the  force  of  attraction 
to  be  yr~n,  where  7  is  a  constant  and  r  is  the  distance  of  the  particle  from 
the  center  of  attraction,  show  that  the  time  taken  to  cover  the  distance 
between  the  point  of  projection  and  the  center  of  attraction  is 


n+lV     27 


MOTION  OF  A   PARTICLE  139 

21.  From  the  following  data  show  that  the  velocity  with  which  a  body 
has  to  be  projected  from  the  moon  in  order  to  reach  the  earth  is  about  1.5 

miles  per  sec.    Both  the  earth  and  the  union  are  supposed  to  be 
The  mass  of  the  moon  is  sV  of  that  of  the  earth.    The  radii  of  the  earth 
ami  the  moon  are  4000  miles  and  lion  miles,  respectively.    The  distance 
between  the  earth  and  the  moon  is  240,000  miles. 

22.  Taking  the  data  of  the  preceding  problem,  show  that  if  the  earth 
and  the  moon  were  reduced  to  rest  they  would  meet,  under  their  mutual 
attraction,  in  about  4.5  days. 


CHAPTER   VII. 
CENTER    OF   MASS   AND    MOMENT    OF   INERTIA. 

CENTER    OF    MASS. 

There  are  two  useful  conceptions,  known  as  center  of 
mass  and  moment  of  inertia,  which  greatly  simplify  dis- 
cussions of  the  motion  of  rigid  bodies.  It  is,  therefore, 
desirable  to  become  familiar  with  these  conceptions  before 
taking  up  the  motion  of  rigid  bodies. 

114.  Definition  of  Center  of  Mass.  —  The  center  of  mass  of 
a  system  of  equal  particles  is  their  average  position;  in  other 
words,  it  is  that  point  whose  distance  from  any  fixed  plane 
is  the  average  of  the  distances  of  all  the  particles  of  the 
system. 

Let  X\,  Xi,  xz,  .  .  .  xn  denote  the  distances  of  the  particles 
of  a  system  from  the  2/2-plane;  then,  by  the  above  definition, 
the  distance  of  the  center  of  mass  from  the  same  plane  is 

-  _  xx  +  x2  +  xz  +  •  •  •    +  xn 
n 

n 

When  the  particles  have  different  masses  their  distances 
must  be  weighted,  that  is,  the  distance  of  each  particle  must 
be  multiplied  by  the  mass  of  the  particle  before  taking  the 
average.  In  this  case  the  distance  of  the  center  of  mass 
from  the  7/2-plane  is  defined  by  the  following  equation: 

(mi+wi*+  •  •  •   +  mn)  x  =  niix1+ m2x2  +  •  •  •   +  mnxn, 

140 


CENTER  OF  MASS  AND   MOMENT  OF   l\i:i;il\       111 


or 

similarly 

and 


*"  Urn  ' 
y      2m' 

Z      I'm  ' 


Evidently  x,  y,  and  z  are  the  coordinates  of  the  center  of 
mass. 

ILLUSTRATIVE    EXAMPLES 

1.  Find  the  center  of  mass  of  two  particles  of  masses  m  and  nm,  which 
are  separated  by  a  distance  a. 

Taking  the  origin  of  the  axes  at  the  particle  which  has  the  mass  m, 
Fig.  72,  and  taking  as  the  .r-axis  the  line  which  joins  the  two  particles  we  get 

0  +  nma 

x  =  — ■ 

m  +  ttm 

7} 

v  =  o, 

5  =  0. 


Y 

_— 

m 
^- — 

c          nm 

U 

-v 

* 

na 
n+1 

n+1 

Fig.  72. 


2.   Find  the  center  of  mass  of  three  particles  of  masses  m,  2  m,  and 
3  m,  which  are  at  the  vertices  of  an  equilateral  triangle  of  si 
Choosing  the  axes  as  shown  in  Fig.  73  we  have 
_  =  0  +  2  ma  +  3  ma  cos  00° 
m  +  2  m  +  3  m 
=   iT:  ". 

_      0  +  0  +  3masjnl 


6  in 


=  1  v  3  o, 

2  =  0. 


142 


A X  A  L VTICAL  MEC HAXICS 


115.  Center  of  Mass  of  Continuous  Bodies. — When  the  par- 
ticles form  a  continuous  body  we  can  replace  the  summation 
si^ns  of  equation  (I')  by  integration  signs  and  obtain  the 
following  expressions  for  the  coordinates  of  the  center  of 
mass: 

rxdm  * 

I dm 

y  dm 
I    dm 

Jz  dm 

*  nm  ' 

/    dm 

Jo 

where  m  is  the  mass  of  the  body. 

ILLUSTRATIVE   EXAMPLES. 

1.  Find  the  center  of  mass  of  the  parabolic  lamina  bounded  by  the 
curves  //'-  =  2  px  and  x  =  a,  Fig.  74. 

Obviously  the  center  of  mass  lies  on  the  x-axis.    Therefore  we  need  to 


(I) 


Fio.  74. 
*  In  general  if  y  is  a  function  of  x  then  the  average  value  of  y  between  the 

1  /•*! 

limits  j-i  and  Jj  is  given  by  the  relation:  y  = I     ydx.' 

Xi  —  X[  J i. 


CENTER  OF  .MASS  AND   MOMENT  OF   [NERTIA       L43 

determine  x  only.    Taking  a  snip  of  width  dx  for  the  element  of  m 

have 

dm  =  a  2  y  dx 

=  2aV'2  pxdx, 

where  a-  is  the  mass  per  unit  area.    Therefore  substituting  this  expression 
of  dm  in  equation  (I)  and  changing  the  limits  of  integration  we  obtain 

Xo            . 
x  \  2  px  dx 
x= -— 

2a  faV~2pxdx 

j   x?  dx 

5  ' 

2.   Find  the  center  of  mass  of  the  lamina  bounded  by  the  curves 
y1  =  4  ox  and  y  =  bx,  Fig.  75. 

Let  dx  dij  be  the  area  of  the  element  of  mass,  then 

dm  =  udxdy. 

Therefore  substituting  in  equation  (I)  and  introducing  the  proper  limits 
of  integration  we  obtain 

4a  '•  ■' 

X,xd'Jdx  -    X  l,'"1"'1* 

X  =  —T^ y  =  —*a 

0       L       dlJdX  X      Xr       ** 

f  bl  (2  V^r"  -  fa)  X  dx  J;"  (^  OX  -  |  X»J  </x 

=         To  ££ 

f  "  (2  V^  -  bx)  dx  fbi(2Va~c-  bx)  dx 

Jo  Jo 

8a 

562'  b 


144 


ANALYTICAL  MECHANICS 


3.   Find  the  center  of  mass  of  a  semicircular  lamina. 
Selecting  the  coordinates  and  the  element  of  mass  as  shown  in  Fig. 
we  have 

dm  =  a  -pdd  -dp, 

C  ("y  -  <?p  dp  de 

-  Jo    .'0 


y  = 


jwj"cxpdpdd 

Cfap-smddpdd 

naP  dp  dd 


dp  fL 

^dm    \ 

1                  a 

ft 

y           \ 

4a 
37T1 
0. 


PROBLEMS. 


Find  the  center  of  mass  of  the  lamina  bounded  by  the  following  curves : 

(1)  y  =  mx,  y  =  —  nix,  and  y  =  a. 

(2)  y  =  a  sin  x,  y  =  0,  x  =  0,  and  x  =  tt. 

(3)  y2  =  ax  and  x2  =  by. 

(4)  x2  +  y2  =  a2,  x  =  0,  and  y  =  0. 

(5)  b2x2  +  a2y2  =  a2b2,  x  =  0,  and  y  =  0. 

(6)  r  =  a{\  +cos0). 

(7)  r  =  a,  6  =  0,  and  6  =  9. 

(8)  r  =  a,r  =  b,d  =  0,  and0=^- 

116.  Center  of  Mass  of  a  Homogeneous  Solid  of  Revolution. 
—  Let  Fig.  77  represent  any  solid  obtained  by  revolving 
a  plane  curve  about  the  .r-axis.  Then  the  center  of  mass 
lies  on  the  axis  of  revolution.  The  position  of  the  center 
of  mass  is  found  most  conveniently  when  the  element  of 
mass  is  a  thin  slice  obtained  by  two  transverse  sections. 
Tlir  expression  for  the  mass  of  such  an  element  is 


dm 


■y-  -  dx, 


where  r  is  the  density  of  the  solid,  y  the  radius  of  the  slice, 
and  '/.'•  its  thickness. 


CENTER  OF  MASS  AND  MOMENT  01    INERTIA      L45 
Y 


Fig. 


ILLUSTRATIVE  EXAMPLE 


Find  the  center  of  mass  of  a  parab- 
oloid of  revolution  obtained  by  revolv- 
ing about  the  x-axis  that  part  of  the 
parabola  y2  =  2  px  which  lies  between 
the  lines  x  =  0  and  x  =  a. 

dm  =  T7T//-  dx 
=  T7r  2  px  dx; 


2tttpC 


..'•-  dx 


2irrp  (    xdx 

=  %a. 

PROBLEMS. 

Find  the  center  of  mass  of  the  homogeneous  Bolid  of  revolution  gener- 
ated by  revolving  about  the  .r-axis  the  area  bounded  by 

(1)  y  =  -x,  x  =  h,  and  y  =  0. 

h 

(2)  x2  =  4ay,  x  =  0,  and  y  =  a. 

(3)  x-  +  y-  =  a2,  and  x  =  0. 

(4)  b2x2  +  <ry-  =  a*b*}  and  x  =  0. 

(5)  y  =  sinx,  x  =  0,  and  x  =  -■ 

(6)  x2  +  >/2  =  0s,  x-  +  //-  -  //-,  and  I  -  0. 


146 


ANALYTICAL  MECHANICS 


117.  Center  of  Mass  of  Filaments.  —  The  transverse  dimen- 
sions of  a  filament  are  supposed  to  be  negligible;  therefore 
it  can  be  treated  as  a  geometrical  curve.  Taking  a  piece  of 
length  ds  as  the  element  of  mass  and  denoting  the  mass  per 
unit  length  by  X  we  have 

dm  =  X  ds. 


ILLUSTRATIVE  EXAMPLE. 

Find  the  center  of  mass  of  a  semicircular  filament. 

(a)  Taking  x2  +  y2  =  a2  to  be  the  equation  of  the  circle  we  get 

dm  =  \ils 


=  x« 

Jo      Va2  -  .r2 

/•a  dx 

Jo  Va2  -  x2 


dx 

Va2  -  x2 
dx 


-V^2 


sin-1  - 
a 


Fig.  79. 


2a 


(b)  Referring  the  circle  to  polar  coordinates  we  have  r  =  a  for  its 
equation.     Therefore 

dm  =  \ds 
=  \ad0. 


f. 


x  Xa  dd 


r. 


a  cos  6  dd 


r 


X  a  ,10 


/ 


dB 


CENTER  OF  MASS  AND   MOMENT  OF   IXKkTIA       \\] 


PROBLEMS. 

Find  the  center  of  mass  of  a  uniform  wire  bent  into  the  following 

curves: 

(1)  An  arc  of  a  circle  subtending  an  angle  2  0  at  the  center. 

(2)  y   =  a  sin  x,  between  x  =  0  and  x  =  ir. 

(3)  y-  =  -i  ax,  between  x  =  0  and  x  =  2  a. 

(4)  The  cycloid  x  =  a  (d  -  sin  0),  ij  =  a  (1  —  cos  6),  between  two 
successive  cusps. 

(5)  Half  of  the  cardioid  r  =  a  (1  +  cos  6). 

118.  Center  of  Mass  of  a  Body  of  Any  Shape  and  Distribution 
of  Mass.  —  The  illustrative  examples  of  the  last  few  pages 
are  worked  out  by  special  methods  in  order  to  bring  out  tin- 
fact  that  in  a  great  number  of  problems  the  ease  with  which 
the  center  of  mass  may  be  determined  depends  upon  the 
choice  of  the  element  of  mass.  The  following  general  ex- 
pressions for  an  element  of  mass  may  be  used  whatever  the 
shape  of  the  body  or  the  distribution  of  its  mass: 

(a)  When  the  bounding  surfaces  of  the  body  are  given 
in  the  Cartesian  coordinates  the  mass  of  an  infinitesimal 
cube   is   taken  as  the  ele- 
ment of  mass : 

dm  =  t  •  dxdy  dz. 

(b)  When  the  bounding 
surfaces  of  the  body  are 
given  in  spherical  coordi- 
nates the  element  of  mass 
is  chosen  as  shown  in  Fig. 
80.  In  this  case  the  fol- 
lowing is  the  expression  for 
the  element  of  mass : 
dm  =  t  •  p  dd  •  dp  •  p  d<f>  sin  0 

=  tp2  sin  d  dd  d<p  dp. 

(c)  When  the  density.  -,  varies  from  point  to  poinl  it 
is  expressed  in  terms  of  the  coordinates  ami  substituted  in 
the  expression  for  dm. 


148 


ANALYTICAL  MECHANICS 


ILLUSTRATIVE    EXAMPLES. 

1.   Find  the  center  of  mass  of  an  octant  of  a  homogeneous  sphere, 
(a)  Suppose  the  bounding  surfaces  to  be 

x2  +  if  +  z*-  =  a-,  x  =  0,  y  =  0,  and  z  =  0. 
Then  the  limits  of  integration  are 

x  =  0  and  x  =  a. 


Therefore 


y  =  0  and  y  =  Vo*  —  x1, 
z  =  0  and  z  =  y/a-  —  x2  —  \ 

j      x  c/x  d#  dz 

j  _  ^o    •'o Jo 

f    f  f    dxdyda 

•'O      •'O  -^  0 

=  3a  * 

8  ' 


and  by  symmetry  y  =  z  =  '—  ■ 

8 

(b)  Suppose  the  equations  of  the 
bounding  surfaces  to  be  given  in  spheri- 
cal coordinates,  then  we  have 


- ,   0  =  0,  and   <t>  =  - 


The  limits  of  integration  are 

r  =  0      and     r  =  a 
0  =  0      and     6  =  ? 


Fig.  81. 


0      and 


C  C   f  V  sin2  0  cos0  rir  rid  d<j> 
Therefore        x  =     °x    ny  '  ° [x  =  r  sin  6  cos0] 

f   f"  C"r-s'mdrirridri<p 

Jo    Jo    Jo 

8  ' 


CENTER   OF   MASS   AND   MOMENT   OF   INERTIA      149 


2.  Find  the  center  of  mass  of  a  righl  circular  cone  whose  density 
varies  inversely  as  the  square  of  the  distance  from  the  apex,  the  distance 

being  measured  along  the  axis. 

dm  —  t  •  iry2  •  dx 

T\  n-.r'1 


Ti7T<7- 


dx, 


where  n  is  the  density  at  a  unit 
distance  from  the  apex.  -There- 
fore 


Tjira-   r' 
hi    J . 


as?  f  c/x 

h 
2 


Fig.  82. 


PROBLEMS. 

1.  Find  the  center  of  mass  of  aright  circular  cone,  the  density  of  which 
varies  inversely  as  the  distance  from  the  vertex. 

2.  Find  the  center  of  mass  of  a  circular  plate,  the  density  of  which 
varies  as  the  distance  from  a  point  on  the  circumference. 

3.  Find  the  center  of  mass  of  a  cylinder,  the  density  of  which  varies 
with  the  nth  power  of  the  distance  from  one  base. 

4.  Find  the  center  of  mass  of  a  quadranl  of  an  ellipsoid. 

5.  Find  the  center  of  mass  of  a  hemisphere,  the  density  of  which  varies 
as  the  distance  from  the  center. 


119.  Center  of  Mass  of  a  Number  of  Bodies.  Lei  mi,  rn:, 
etc.,  be  the  masses  and  xu  .<..  etc.,  be  the  x-coordinates  of  tin- 
centers  of  mass  of  the  individual  bodies.  Then  if  x  denotes 
the  z-coordinate  of  the  center  of  ma—  of  the  entire  system 
we  can  write 


x  = 


xdm 


150 


Similarly 


ANALYTICAL  MECHANICS 

JfV»i                         f™2 
xdm-\-  I     xdm+  ■ 
o Jo 

dm+   I     dm+  ■  • 

o  Jo 

_  m&\  +  m2^2  +  •  •  • 

mi  +  m2+  •  •  • 
_  HmXj 
2m 


2m 

1/// 


Therefore  the  mass  of  each  body  may  be  considered  as  being 
concentrated  at  its  center  of  mass. 

ILLUSTRATIVE   EXAMPLE. 

Find  the  center  of  mass  of  the  plate  indicated  by  the  shaded  part  of 
Fig.  83. 

(a)  Suppose  the  plate  to  be  separated 
into  two  parts  by  the  dotted  line.  Then 
the  coordinates  of  the  center  of  mass  of 
the  lmver  part  are 


b  ,     -       b  —  a 

Xi  =  -     and     i/i  = 


2      "  2 

On  the  other  hand  the  coordinates  of  the 
cciiicr  of  mass  of  the  upper  part  are 

2b  -a 


b  —  a 
'  2 


and    y2  = 


Therefore  the  coordinates  of  the  center  of  mass  of  the  entire  plate  have 
the  following  values: 


b  b  —  a 

»"2  +  w*  — 

mi  +  m» 
oh(h-a)-+<ra(b  —  a)  — ^ 

ab  (6— o)+ff  (b  —  a)  a 

b-  +  nb  -  q' 
2(a  +  6) 


2 


+  m, 


26 


TO]  +  I »  2 

j  n        .b  —  a.       ,,       ,2  b  — a 
ab{b-a)— — \-aa(b-a)  — — 

ab  (b  —  a)-\-aa  {b  —  a) 
b-  +  nb  -  a2 
2  (a +  6) 


CENTER  OF  MASS  AND  MOMENT  OF   IM.bTIA       151 

(b)  Suppose  the  square  OA  to  represenl  a  plate  of  positive  ma 
the  square  O'A  to  represent  a  plate  of  negative  mass.  Then  if  the  two 
plates  have  the  same  thickness  and  density  the  positive  and  the  m 
masses  annul  each  other  in  the  square  O'A.  Therefore  the  two  Bquare 
plates  form  a  system  which  is  equivalent  to  the  actual  plate  represented 
by  the  shaded  area  of  the  figure.  Bence  I  he  center  of  mass  of  the  Bquare 
plates  is  also  the  center  of  mass  of  the  given  plate. 

The  masses  of  the  square  plates  are  air  and  -  an-,  while  the  codrdi- 
nates  of  their  centers  of  mass  are 

-,      -,     b         ,    -„     -„     26 -a 

x  =  y  =  -     and    x   =  y   =  — - —  • 

Therefore  the  coordinates  of  the  center  of  mass  of  the  two  are 

,„&   .    ,          „.  2b-a 
ffb*-+  {-an-)  — 


ah-  +  (  —  ad1) 
b-  -\-  ab  —  a- 


2  (a  +  b) 
which  are  identical  with  those  obtained  by  the  first  method. 


PROBLEMS. 
1.   Find  the  center  of  mass  of  the  homogeneous  plates  indicated  by  the 
following  figures: 


V'aJ 

■\ 

/aa\ 

2a 
a 

a        al 

(a) 


(b) 


(r) 


2.  A  s])liere  of  radius  6  has  a  spherical  cavity  of  radius  "•  find  the 
center  of  mass  if  the  distance  between  the  center 

3.  A  right  cone  is  cut  from  a  righl  circular  cylinder  of  the  same  base 
and  altitude.     Find  the  center  of  mass  of  the  remaining  rolid. 

4.  A  right  cone  is  cut  from  a  hemisphere  of  the  sane  base  and  altitude. 
Find  the  center  of  mass  of  the  remaining  solid. 

5.  A  right  circular  cone  is  cu1  from  another  right  circular  cone  of  the 
same  base  but  of  greater  altitude,  bind  the  center  of  mass  of  t  he  remain- 
ing solid. 

6.  A  right  circular  cone  is  cul  fn»m  the  paraboloid  of  revolution  genfir- 


152 


ANALYTICAL   MECHANICS 


ated  by  revolving  about  the  x-axis  the  area  bounded  by  y2  =  2  px 
and  x  =  a.  Find  the  center  of  mass  of  the  remaining  solid  if  the  parab- 
oloid and  the  cone  have  the  same  base  and  vertex. 

MOMENT  OF  INERTIA. 
120.  Definition  of  Moment  of  Inertia. — The  moment  of 
inertia  of  a  body  about  an  axis  equals  the  sum  of  the 
products  of  the  masses  of  the  particles  of  the  body  by  the 
square  of  their  distances  from  the  axis.*  Thus  if  dm  de- 
notes an  element  of  mass  of  the  body  and  r  its  distance 
from  the  axis  then  the  following  is  the  analytical  state- 
ment of  the  definition  of  moment  of  inertia: 

/  =  f>  dm.  (II) 

The  integration  which  is  involved  in  equation  (II)  is  often 
simplified  by  a  proper  choice  of  the  element  of  mass.  The 
choice  depends  upon  the  bounding  surfaces  of  the  body  and 
the  position  of  the  axis ;  therefore  there  is  no  general  rule  by 
which  the  most  convenient  element  of  mass  may  be  selected. 
There  is  one  important  point,  however,  which  the  student 
should  always  keep  in  mind  in  selecting  the  element  of  mass, 
namely,  the  distances  of  the  various  parts  of  the  element  of  mass 
from  the  axis  must  not  differ  by  more  than  infinitesimal  lengths. 


ILLUSTRATIVE   EXAMPLES. 
1.    Find  the  moment  of  inertia  of  a  rectangular  lamina  about  one  of  its 

Y 


dm 


_L x 


dy 


sides. 

Suppose  iIh'  lamina  to  lie  in  the  xy- 
plane.  Further  suppose  the  side  with 
reaped  to  which  the  moment  of  inertia  is 
to  lie  found  to  lie  in  the  x-axis.  Then 
the  mOSl    Convenient  element   of    mass   is 

a  strip  which  is  parallel  to  the  x-axis. 

Let  ,/  be  the  length  (Fig.  84),  b  the 
width,  and  <t  the  mass  per  unit  area  of  the  lamina,  then 
dm  =  aady. 

'    I  <>r  :i  physical  definition  of  moment  <>f  inertia  and  its  meaning  see  p.  220. 


Fig.  84. 


CENTER  OF  MASS  AND   MOMENT  OF    INERTIA       153 


The  distance  of  the  element  of  mas-;  from  the  axis  is  y\  therefore  substi- 
tuting in  equation  (II)  these  expressions  for  dm  and  its  distance  from  the 

axis  we  obtain 

I  =   )  //-  •  an  dy 

=  i  (rub3 

=  i  mb2, 

for  the  desired  moment  of  inertia.  The  limits  of  integration  arc  different 
from  those  in  equation  (II)  because  the  independent  variable  is  changed 
from  m  to  y, 

2.  Find  the  moment  of  inertia  about  the  x-axis  of  a  lamina  which  is 
bounded  by  the  parabola  x2  =  2  py  and  the  straight  line  y  =  a. 

(a)  Choosing  a  horizontal  strip  for  the  element  of  mass  we  have 

dm  =  a  -  2xdy. 


/a 
xf-xdy 
0 

=  2  a  f  y2V2py  dy 
J  0 


>'• 


But 


m  =  a  I    2  x  dy 

•J  0 

=  I  a  a  v2  pa. 


(b)  We  can  also  take  an  element  of  the  strip  for  the  element  of  mass, 
in  which  case  we  have 

dm  =  a  dx  dy. 

:.     I  =    \  ir  •  a  dx  dy 

=  i  a  a3  V'2  pa, 
=  2  ma2. 


154 


AXA I ATICAL  MECHANICS 


PROBLEMS. 

1.  Find  the  moment  of  inertia  of  a  circular  lamina  about  a  diameter. 

2.  Find  the  moment  of  inertia  of  an  elliptical  lamina  about  its  minor 
axis. 

3.  Bind  the  moment  of  inertia  of  a  rectangular  plate  of  negligible 
thickness  about  a  diagonal. 

4.  Find  the  moment  of  inertia  of  a  thin  plate,  which  is  in  the  shape  of 
an  equilateral  triangle,  with  respect  to  one  of  its  edges. 

5.  Find  the  moment  of  inertia  of  a  triangular  plate  about  an  axis 
which  i  lasses  through  one  of  its  vertices  and  is  parallel  to  the  base. 

6.  Find  the  moments  of  inertia  of  the  following  lamina  with  respect 
to  the  axes  indicated  by  the  thin  vertical  lines. 


a 
2b 

Ah 

b 

a 
2a  b             ; 

1        1 

J 

b 

1 

n 

3b 
a 

2a 
2b 

a  1 

a 

2b 
2a 

(a)  (b)  (c)  (d)  (e) 

121.  Theorems  on  Moments  of  Inertia.  Theorem  I.  —  The 
moment  of  inertia  of  a  lamina  about  an  axis  which  is  per- 
pendicular to  its  plane  equals  the  sum  of  the  moments  of  inertia 
with  respect  to  two  rectangular  axes  which  lie  in  the  plane  of 
the  lamina  with  their  origin  on  the  first  axis. 

Suppose  the  lamina  to  be  in  the  xy-p\a,ne,  then  the  theo- 
rem states  that  the  moment  of  inertia  about  the  z-axis 
equals  the  sum  of  the  moments  of  inertia  about  the  other 
two  axes,  that  is, 

It  =  I*+Iv.  (Ill) 

The  following  analysis  explains  itself. 

{TO 
r2dm 

(x2+  y'1)  dm 

xidm+  J    y*dm 


1 


CENTER  OF   MASS   AND    MOMENT  OF    INERTIA       155 


It  is  evident  from  this  theorem  thai   when  the  lamina  is 
rotated  about  the  z-axis  Iz  and  Iy  change,  in  general,  but 

their  sum  remains  constant. 

122.  Theorem  II.  —  The  moment  of  inertia  of  a  body  about 

any  axis  equals  its  moment  of  inertia  about  a  parallel  axis 

through  the  center  of  7nass 

plus  the  product  of  the  mass 

of  the  body  by  the  square  of 

the  distance  between  the  two 

axes. 

Let  the  axis  be  perpen- 
dicular to  the  plane  of  the 
paper  and  pass  through 
the  point  0,  Fig.  86.  Fur- 
ther let  dm  be  any  ele- 
ment of  mass,  r  its  dis- 
tance from  the  axis  through  0,  and  rc  its  distance  from  a 
parallel  axis  through  the  center  of  mass,  C.  Then  if  a 
denotes  the  distance  between  the  axes  we  have 


Fig.  S6. 


dm 


/=/   r»« 

Jo 
=  I    (r2  +  a2  —  2  arc  cos  0)  dm 

rc-dm+\    a* dm— 2a  \    rc cos d dm 

0  i/O  t/0 

=  Ic+  ma2-  2a  j   xdm. 

nm 

But  by  the  definition  of  the  center  of  mass    J    xdm=  tnx, 

and  in  the  present  case  the  center  of  mass  is  at  the  origin; 
therefore  x  and  consequently   the   lasl    integral   \ranishes. 

Thus  we  get 

l  =  I.+  ma2.  IV) 

123.  Radius  of  Gyration. — The  radius  of  gyration  of  a  body 

with  respect  to  an  axis  is  defined  as  the  distance  from  the 


156 


ANALYTICAL  MECHANICS 


axis  of  a  point  where  if  all  the  mass  of  the  body  were  con- 
centrated its  moment  of  inertia  would  not  change. 

Lei  m  denote  the  mass  of  a  body,  /  its  moment  of  inertia 
with  respect  to  a  given  axis,  and  K  its  radius  of  gyration 
relative  to  the  same  axis;  then  the  definition  gives 
/  =  K2m,  ) 

K=\fl]  (V) 

'  m  J 

If  Ke  denote  the  radius  of  gyration  relative  to  a  parallel  axis 
through  the  center  of  mass,  then  by  equations  (IV)  and  (V) 
we  obtain 

K2=K2  +  a2.  (VI) 


ILLUSTRATIVE   EXAMPLES. 

1.   Find  the  moment  of  inertia  of  a  homogeneous  circular  disk  (a)  about 
its  geometrical  axis,  (b)  about  one  of  the  elements  of  its  lateral  surface. 

Let  m  be  the  mass,  a  the  radius,  I  the  thickness,  and  r  the  density  of  the 
disk.    Then  choosing  a  circular  ring  for  the  element  of  mass  we  have 
dm  =  t  •  /  •  2  xr  -dr, 

where  r  is  the  radius  of  the  ring  and  dr  its 
thickness.  Therefore  the  moment  of  inertia 
about  the  axis  of  the  disk  is 

I  =  2ttIt  C"r*dr 

Jo 
_  rlira* 
2 

_  mo2 

2   *  Fig.  87. 

The  moment  of  inertia  about  the  element  is  obtained  easily  by  the  help 
of  theorem  II.    Thus 

/'  =  /  +  ma* 

=  *  ma2. 

It  will  he  noticed  t li.it  the  1  hiekness  of  the  disk  does  not  enter  into  the 
ione  for  /  and  /'  excepl  through  the  mass  of  the  disk.    Therefore 

.pressiniis  hold  good  whether  the  disk  is  thick  enough  to  be  called 
a  Cylinder  or  thin  enough  to  he  called  a  circular  lamina. 


CENTER  OF  MASS  AND  MOMENT  OF   [NERTIA       157 

2.   Find  the  moment  of  inertia  of  a  cylinder  aboul  a  transverse  axis 
through  the  center  of  mass. 

Let  m,  a,  I,  and  r  be,  respectively,  the  mass,  the  radius,  the  Length,  and 
the  density  of  the  cylinder.     Further  lei  the  given  axis  pass  through  the 

center  of  mass  of  the  cylinder;  then  taking  a  slice  obtained  by  two  right 
sections  as  the  element  of  mass  we  get,  by  theorem  II. 

(II  y     =     <lly>     +    Z-    <llll, 

where  dm  is  the  mass  of  the  element,  dly  and  dly-  are  the  momenta  of 
inertia  of  the  element  aboul  the  given  axis  and  aboul  a  parallel  axis  through 

the  center  of  mass  of  the  element,  and  r  is  the  distance  In 'tween  th 
axes.     But  by  theorem  I 

ill*  +  die  =  dl» 

and  by  symmetry  dl^  =  dl^, 

and  by  the  last  illustrative  example 

a2  dm 
•> 


dls  = 


Therefore 


dlj 


a-  dm 


F^l 


Vu,.  88. 


Substituting  this  value  of  dlu   in  the  expression  for  <//„  we  gel 
dly  =  (^  +  :^dm 

Integrating  the  last  equation  we  have 

y        ma-   ,     rm    •  i 
/„  =  — +J     z*dm 

i 
mn'1    ,    r  -  .,  .,  , 


158 


ANALYTICAL  MECHANICS 


—  m°2  _l_  Tira~l3 

4    +     12 

3.   Find  the  moment  of  inertia  of  a  homogeneous  sphere  about  a 
diameter  ami  about  a  tangent  line. 

Let  )n,  a,  and  r  be  the  mass,  the    y 
radius,  and  the  density  of  the  sphere, 
respectively.     Then,   taking   the  axes 
and  the  element  of  mass  as  shown  in 
Fig.  89,  we  have 

dly  =  dly»  +  22  dm, 

=  2 

_  1  y-  dm 
~  2     2 
Integrating  the  last  equation 


dIM  +  z2  dm 

+  z2dm. 


Iv=ifQ  !J2<l>»+f0   ^dm 

—  "7    I    y4dz  -\-ttt  I    z2y2  dz  [dm  =  mry-dz] 

4    J— a  J  —a 

= f  r  ^  -  ^2  dz + 2  ^  i>2  -  *>  ■*■ 


and 


Sttto5 
15 

=  I  TOO2 

Iy>  =  Iy  +  ma- 
=  I  ma2. 


124.  Theorem  III.  —  77/ r  moment  of  inertia  of  a  homogeneous 
right  cylinder  about  a  transverse  axis  equals  the  moment  of  in- 
ertia of  two  lamince  which  fulfill  the  following  conditions,  (a) 
Each  lamina  has  a  mass  equal  to  that  of  the  cylinder.  (6)  One 
lamina  occupies  the  entire  area  of  the  transverse  section  of  the 
cylinder  through  the  given  axis,  while  the  other  lamina  occupies 
the  <  ntire  area  of  the  longitudinal  section  of  the  cylinder  through 
(he  "j    . 


CENTER  OF  MASS  AND    MOMENT  OF    [NERTIA       159 

Let  Y,  Fig;.  00,  be  the  axis  with  respect  to  which  it  is 
desired  to  find  the  moment  of  inertia  of  the  cylinder.  Lei 
dlv denote  the  moment  of  inertia  of  an  element  bounded  by 
two  transverse  sections  relative  to  1  he  K-axis,  and  ///,..  denote 
the  moment  of  inertia  of  the  same  element  relative  to  the 


Fig.  90. 

F'-axis,  a  parallel  axis  through  the  center  of  mass  of  the 
element.     Then,  by  theorem  II,  we  have 
dly=  dly»+  (x2+z2)clm, 

where  (x2+z2)  is  the  square  of  the  distance  between  tin- 
two  axes.  Similarly  the  moment  of  inertia  of  the  element 
about  the  F'-axis  which  is  parallel  to  the  F"-axis  and  inter- 
sects the  same  elements  of  the  cylinder,  is  given  by 

dlv>=  dlv-  +  z2dm. 
Eliminating  dlv-  between  the  last  two  equations  we  obtain 

dly=  dly-  +  x*dm 
=  Ki'-d/n  +  x-drn, 


160 


ANALYTICAL  MECHANICS 


where  A'i  is  the  radius  of  gyration  of  the  element  of  mass 
about  the  F'-axis.     Integrating  the  last  equation  we  have 

Iv=  I   Kx*dm+   I   x-dm. 

Jo  Jo 

Each  of  the  elements  of  mass  has  its  own  F'-axis  similarly 
placed.  Therefore  ki  is  the  same  for  all  the  elements  of  mass 
and  remains  constant  during  the  integration.     Hence 

Iv  =  Ki2m+   I   x-dm 

where  h  =  Kx-m  and  72  =    /    #2  dm.     It  is  not  difficult  to  see 

that  h  is  the  sum  of  the  moments  of  inertia  of  all  the  elements 
of  mass  relative  to  their  respective  F'-axes.  It  is  equal, 
therefore,  to  the  moment  of  inertia  about  the  F-axis  of  the 
lamina  (A  in  the  figure)  which  would  be  obtained  if  the  en- 
tire cylinder  could  be  compressed  into  the  transverse  section 
through  the  F-axis.  On  the  other  hand  72  equals  the  mo- 
ment of  inertia  about  the  F-axis  of  the  lamina  (D  in  the 
figure)  which  would  be  obtained  if  the  cylinder  could  be  com- 
pressed into  the  longitudinal  section  through  the  F-axis. 


ILLUSTRATIVE  EXAMPLE. 
As  an  illustration  of  the  last  theorem  consider  the  illustrative  example 


of  p.  157. 


( 

■             = 

>ft^ 

Sil 

K\ 

Fig.  91. 

Applying  the  theorem  we  sec  thai  the  moment  of  inertia  of  the  cylin- 
der equals  the  sum  of  the  momenta  <»f  inertia  of  the  two  lamina'  of  Fig.  91. 


CENTER  OF  MASS  AND  MOMENT  OF  ENERTIA      Hil 

Denoting  the  moment  of  inertia  of  the  circular  lamina  by  I \  and  that 
of  the  rectangular  lamina  by  h  we  have 

/  =  /i  +  u 

I 
1  ma*  .    r 5    »  j 


2  /-2 

-  +  era  I    j  x2  dr 

_ 2 

Z2 


<?+§> 


which  is  identical  with  the  result  obtained  by  the  direct  method. 

PROBLEMS. 

1.  Find  the  moment  of  inertia  of  a  hollow  circular  cylinder  with  re- 
spect to,  (a)  its  geometrical  axis,  (b)  an  element,  (c)  a  transverse  axis 
through  the  center  of  mass. 

2.  Find  the  moment  of  inertia  of  an  elliptical  cylinder  with  respect  to, 
(a)  its  geometrical  axis,  (b)  a  transverse  axis  through  its  center  of  mass 
and  parallel  to  the  major  axis  of  a  right  section. 

3.  Find  the  moment  of  inertia  of  a  rectangular  prism  with  respect  to, 
(a)  its  geometrical  axis,  (b)  a  transverse  axis  through  the  center  of  mass 
and  perpendicular  to  one  of  its  faces. 

4.  In  the  preceding  problem  suppose  the  prism  to  be  hollow. 

5.  Find  the  moment  of  inertia  of  a  prism,  the  cross  section  of  which  is 
an  equilateral  triangle,  with  respect  to,  (a)  its  geometrical  axis,  do  a 
transverse  axis  through  its  center  of  mass  and  perpendicular  to  one  of  its 
faces. 

6.  In  the  preceding  problem  suppose  the  prism  to  be  hollow. 

7.  Find  the  moment  of  inertia  of  a  hollow  sphere  with  respect  to,  (a) 
a  diameter,  (b)  a  tangent  line. 

8.  Find  the  moment  of  inertia  (if  a  spherical  shell  of  negligible  thick- 
ness with  respect  to  a  tangent  line. 

9.  Find  the  moment  of  inertia  of  a  right  circular  cone  with  reaped  to, 
(a)  its  geometrical  axis,  (b)  a  transverse  axis  through  the  vertex. 

10.  In  the  preceding  problem  suppose  the  cone  to  be  a  -hell  of  negli- 
gible thickness. 

11.  Find  the  moment  of  inertia  of  a  paraboloid  of  revolution  with 
respect  to,  (a  I  its  axis,  (b)  a  transverse  axis  through  its  vertex.    The  radius 

of  the  base  is  a  and  the  height  is  h. 


162  ANALYTICAL  .MECHANICS 

12.  Find  the  moment  of  inertia  of  an  ellipsoid  with  respect  to,  (a)  one 
of  its  axes,  (b)  a  tangent  at  one  end  of  one  of  the  axes  parallel  to  the 
other. 

13.  In  the  preceding  problem  suppose  the  body  to  be  an  ellipsoidal 
shell  of  negligible  thickness. 

125.  General  Method.  —  The  special  methods  which  have 
been  discussed  in  the  last  few  pages  are  desirable  but  not 
necessary  for  finding  the  moments  of  inertia  of  bodies.  In- 
stead of  selecting  special  types  of  elements  of  mass  for  each 
type  of  bodies  and  then  making  use  of  the  various  theorems 
we  can  use  the  general  expressions  for  dm  which  were  given 
on  p.  147  and  obtain  the  moment  of  inertia  directly  from 
equation  (II). 

As  an  illustration  of  this  general  method  consider  the  moment  of  inertia 
of  a  sphere  with  respect  to  a  diameter.  It  is  evident  from  the  symmetry 
of  the  body  that  the  moment  of  inertia  about  a  diameter  equals  eight 
times  that  of  an  octant  about  one  of  its  straight  edges. 

(a)  Let  the  octant  be  taken  as  shown  in  Fig.  81,  and  be  referred  to 
Cartesian  coordinates,  then  the  equations  of  the  bounding  surfaces  are 

x2  +  if  +  z2  =  a2,     x  =  0,  y  =  0,  and  z  =  0. 
Hence  taking  the  x-axis  as  the  axis  of  reference  we  have 
/  =  Jmr2  d)n 

=  8rJoJo         Jo    W  +  '^dxdydz 
=  I  ma2. 

(b)  Let  the  octant  be  referred  to  spherical  coordinates,  Fig.  80,  then 
the  equations  of  the  bounding  surfaces  are 

and  p 
Therefore  1  =  f 


TV 

~  2 

,  4>  = 

0,   0 

TV 

dm 

2  - 

p2sin 

0COS2 

4>)  dm 

J  0 

8t  ('  P  f'p*  (sin  d  -  sin3  6  cos24>)  dd  d<p  dp 
Jo  J o Jo 


CENTER  OV   MASS   AM)    MOMENT  OF   [NERTTA      1G3 

126.  Routh's  Rule.  —  The  following  is  a  useful  rule  for  re- 
membering the  moments  of  inertia  of  certain  types  of  bodies: 
Moment  of  inertia  with  respect  to  any  axis  of  symmetry 

.  .  sum  of  the  squares  of  the  perpendicular  semi-axes 

=  mass  X J- ^—        J         H    y  -. 

8,  4,  5 

The  denominator  of  the  right-hand  member  is  3,  4,  or  5  ac- 
cording as  the  body  is  rectangular,  elliptical,  or  ellipsoidal. 

The  following  illustrate  Routh's  rule. 

Rectangular  lamina;  about  axis  perpendicular  to  its  plane: 

«!  +  fr; 

.  4       4  a2  +  b2 

Circular  lamina;  about  axis  perpendicular  to  its  plane: 

/      m  a2  +  a?  _  ma2 
J  =  m_____. 

Elliptical  lamina;  about  axis  perpendicular  to  its  plane: 

T  a2  +  b2 

I  =  m — 

4 

Rectangular  parallelopiped;  about  axis  perpendicular  to  one  of  its  sides : 


I  =  m 


4       4       a2  4-  b2 


3  12 

Circular  cjdinder;  about  longitudinal  axis: 
r  a2  +  a2      ma2 

Sphere;  about  a  diameter: 

o-  +  a2       2 


_  ma2. 
5 


Ellipsoid;  about  one  of  its  axe-: 

.          a2  4-  b2 
I  =  m — - 


CHAPTER   VIII. 
WORK. 

127.  Work.  —  The  mechanical  result  produced  by  the  ac- 
tion of  a  force  in  displacing  a  particle  may  be  considered  to 
be  proportional  to  the  interval  of  time  during  which  the  force 
acts  or  to  the  distance  through  which  it  moves.  In  other 
words,  we  can  take  either  the  time  or  the  displacement  as 
the  standard  of  measure.  The  effect  measured  when  the 
time  is  taken  as  the  standard  is  different  from  that  which 
is  obtained  when  the  displacement  is  made  the  standard. 
The  first  effect  is  called  impulse.  It  will  be  discussed  in  a 
later  chapter.  The  second  is  called  work,  the  subject  of 
this  chapter. 

128.  Measure  of  Work. — When  a  force  moves  a  body  it 
is  said  to  do  work.  The  amount  of  work  done  equals  the 
product  of  the  force  by  the  distance  through  which  the  body 
is  displaced  along  the  line  of  action  of  the  force.  In  this  defi- 
nition the  force  is  considered  to  be  constant.  When  it  is  vari- 
able  the  definition  holds  for  infinitesimal  displacements,  since 
during  the  time  taken  by  an  infinitesimal  displacement  the 
force  may  be  considered  as  constant.  Therefore  if  the  par- 
ticle P,  Fig.  92,  is  displaced  through  ds, 
under  the  action  of  the  force  F,  the  work 
done  is 

dW  =  F  -ds  cos  a, 

where  a  is  the  angle  between  the  direc- 
tions of  the  force  and  the  displacement. 
When  the  displacement  is  finite  the 
work  'lone  equals  the  sum  of  the  amounts  of  work  done  in 

164 


WORK  165 

successive  infinitesimal  displacements.     Therefore  the  work 
done  in  any  displacement  is  given  by  the  integral 


j: 


F cos  ads.  (I) 


When  the  path  of  the  particle  is  curved  the  direction  of  ds 
coincides  with  that  of  the  tangent  to  the  curve.  Therefore 
F  cos  a  is  the  tangential  component  of  the  force.  In  other 
words  the  tangential  component  of  force  does  all  the  work. 
Hence 

PsFT  ds.  (I') 


*-r 


The  normal  component  does  no  work  because  the  particle 
is  not  displaced  along  it. 

Special  Cases.  Case  I. — When  the  force  is  constant,  both 
in  direction  and  in  magnitude,  it  can  be  taken  out  of  the 
integrand.     Therefore 

W=  F  I  cos  a  ds. 

The  last  integral  equals  the  projection  of  the  path  upon  the 
direction  of  the  force.  Therefore  the  product  of  the  force 
by  the  projection  of  the  path  upon  the  line  of  action  of  the 
force  equals  the  work  done. 

Case  II.  —  When  the  force  is  constant  and  the  path  is 
straight  then  the  angle  between  the  force  and  the  displace- 
ment is  constant.     Therefore 

W  =  Fcosa  /  ds 

=  Fs  COS  a. 

Case  III.  —  When  the  force  is  not  only  constant  bul  is 
also  parallel  to  the  path,  then  a  =  0.     Therefore 
W  =  Fs. 

Case  /F.  — When  the  force  is  at  right  angles  to  the  dis- 
placement a  =?|,  and  cos  a  =  0.  Hence  W  0.  Therefore 
the  force  does  no  work  unless  it  has  a  component  along  the 


166 


ANALYTICAL   MECHANICS 


path.  In  this  case  the  motion  of  the  particle  is  not  due  to 
the  force  in  question. 

129.  Work  Done  Against  the  Gravitational  Force. —  These 
special  cases  may  be  illustrated  by  considering  the  work 
done  in  raising  a  body  from  a  lower  to  a  higher  level  against 
the  gravitational  attraction  of  the  earth.  Consider  the 
work  done  in  taking  a  particle  from  A  to  B,  along  each  of 
the  three  paths  shown  in  Fig.  93. 

(a)  Suppose  the  particle  to  be  taken  from  A  to  C  and 
then  to  B;  the  work  done  in  taking  it  from  A  to  C  comes 
under  Case  IV.  The  direction  of  motion  is  at  right  angles 
to  that  of  the  gravitational  force,  therefore  no  work  is  done 
against  it.  The  work  done  in  taking  the  particle  from  C  to 
B  comes  under  Case  III;  the  force  and  the  motion  are  in 
the  same  direction.     Therefore  the  work  done  is 


W 
the 


=  mgh, 
vertical 


height    of   B 


where    h 
above  A. 

(b)  Suppose  the  particle  to  be  taken 
along  the  straight  line  AB.  This  comes 
under  Case  II.  The  angle  between  the 
force  and  the  direction  of  motion  is 
constant.     Therefore 

W  =  mgl  cos  a, 

where  I  is  the  length  of  the  line  AB. 
Bui  since  I  cos  a  =  h,  the  work  done  is 
the  same  as  in  (a),  that  is,  mgh. 

C     Suppose  the  particle  to  be  taken  along  the  curve  AB. 
This  comes  under  ( 'use  I.    Then 


Fia.  93. 


W  =  nig  j  cos  ads 
Jo 

=  mi/  j   <llt  [since  ds  cos  a  =  dh] 

=  mgh. 


WORK-  167 

Therefore  the  work  dom  against  the  gravitational  force  in 
taking  a  body  from  one  position  to  another  depends  only  upon 
the  vertical  height  through  which  the  body  is  raised  and  not 

upon  the  path. 

130.  Dimensions  and  Units  of  Work.  —Work  is  a  scalar 
magnitude  and  has  for  its  dimensions  [ML-T~-\.  The  ( '.<  r.S. 
unit  of  work  is  the  erg.  It  equals  the  work  done  by  a  force 
of  one  dyne  in  displacing  a  particle  through  a  distance  of 
one  centimeter,  measured  along  its  line  of  action.     It   is 

symbolized  by  - — '- — ~.     The  erg  is  a  small  unit,  therefore 
sec.- 

a  larger  unit  called  the  joule  is  also  used. 
1  joule  =  107  ergs. 

The  British  unit  of  work  is  the  foot-pound  (ft. -lb.).  It  is 
the  work  done  against  the  gravitational  attraction  of  the 
earth  in  lifting  one  pound  through  a  vertical  distance  of 
one  foot.  Since  the  work  done  in  lifting  bodies  is  mgJi,  we 
can  express  the  foot-pound  in  terms  of  the  fundamental 
units,  thus 

1  ft.  lb.  =  1  pd.  X  32.2-^r  X  1  ft. 
sec.2 

=  32.2PdJL2, 
sec.2 

where  pd.  represents  the  pound-mass. 

131.  Work  Done  by  Components  of  Force. — The  work  done 
by  a  force  F  in  giving  a  particle  a  displacement  ds  i-  Fcos  6ds} 
where  6  is  the  angle  between  F  and  ds.  Let  X.  Y,  and  Z  be 
the  rectangular  components  of  F,  then  the  direction   of  F 

is  defined  by  its  direction  cosines  — ,  — ,  ami     .     Therefore 

if  /,  m,  and  n  denote  the  direction  cosines  of  ds.  we  get 

V  V         Z* 

cosd=l'F  +m-  +  n- 

*  S<t  Appendix  Arv. 


168  ANALYTICAL  MECHANICS 

and  F  cos  d  ds  =  (IX  +  mY  +  nZ)  ds 

=  Xdx  +  Ydy  +  Z  dy 

where  dx,  dy,  and  dz  are  the  components  of  ds  along  the 
axes.  Thus  the  total  work  done  in  a  finite  displacement  is 
given  by 

W  =  f  F  cos  d  ds 

Jo 

=  f*Xdx+  fyYdy+  Pzdz.  (II) 

Jo  Jo  Jo 

Equation  (II)  states  that  the  work  done  by  a  force  equals 
the  sum  of  the  amounts  of  work  done  by  its  components. 


PROBLEMS. 

1.  Find  the  number  of  foot-pounds  in  one  Joule. 

2.  Find  the  number  of  ergs  in  one  foot-pound. 

3.  Find  the  work  done  in  dragging  a  weight  w  up  an  inclined  plane 
of  length  I,  height  h,  and  coefficient  of  friction  fi. 

4.  A  body  of  100  kg.  mass  is  dragged  up,  then  down,  an  inclined  plane. 
Compare  the  work  done  in  the  two  cases  if  the  length  of  the  plane  is  15  m., 
the  height  5  m.,  and  the  coefficient  of  friction  0.5. 

5.  What  is  the  work  done  in  winding  a  uniform  chain  which  hangs  from 
a  horizontal  cylinder?    The  chain  is  25  in.  long,  and  has  a  mass  of  125  kg. 

6.  A  body  has  to  be  dragged  from  a  point  at  the  base  of  a  conical  hill 
toa  poinl  diametrically  opposite.  Show  that,  if  the  angle  which  the  sides 
of  the  hill  make  with  the  horizon  equals  the  angle  of  friction,  the  work 
done  in  dragging  the  body  over  the  hill  is  less  than  in  dragging  it  around 
the  base. 

7.  A  steam  hammer  falls  vertically  from  a  height  of  3  feet  under  the 
action  of  its  own  weight  and  of  a  force  of  2000  pounds  due  to  steam  pres- 
sure At  the  end  of  its  fall  it  makes  a  dent  of  1  inch  depth  in  an  iron 
plate  Find  the  total  amount,  of  work  done  in  making  the  dent.  The 
hammer  weighs  loot)  pounds. 

8.  In  the  preceding  problem  find  the  average  resisting  force. 

9.  A  locomotive  which  is  capable  of  exerting  a  draw-bar  pull  of  1.5  tons 
is  coupled  to  B  train  of  six  cars.     The  locomotive  and  the  tender  weigh 

The  cars  weigh  15  tons  each.  Find  the  time  it  takes  the  loco- 
motiveto  imparl  to  t lie  train  a  velocity  of  00  miles  per  hour  and  the  work 
done  under  the  following  conditions. 


WORK 


169 


(a)  Horizontal  tracks  and  no  resistance. 

(b)  Horizontal  tracks  and  a  resistance  of  12  pounds  per  ton. 

(c)  Down  a  grade  of  1  in  200  with  no  resistance. 

(d)  Same  as  in  (c)  kit  with  a  resistance  of  l_'  pounds  per  ton. 

(e)  Up  a  grade  of  1  in  200  with  do  resistance. 

(f)  Same  as  in  (b)  but  with  a  resistance  of  l_'  pounds  per  ton. 

10.  A  mass  of  5  pds.  is  at  the  bottom  of  a  vertical  shaft  which  reaches 
the  center  of  the  earth.  How  much  work  will  have  to  be  done  in  order 
to  bring  it  to  the  surface?  The  weight  of  a  body  varies,  within  the  cart  h, 
directly  as  its  distance  from  the  center.  Take  4000  miles  to  be  the  depth 
of  the  shaft. 

11.  Express  the  result  of  the  last  problem  in  joules. 

132.  Work  Done  by  a  Torque.  —  Suppose  the  rigid  body  A, 
Fig.  94,  to  be  given  an  angular  displacement  dd  about  a  fixed 
axis  through  the  point  O,  per- 
pendicular to  the  plane  of  the 
paper.  The  displacement  may 
be  considered  to  be]  due  to 
a  single  force  which  forms  a 
couple  with  the  reaction  of  the 
axis,  or  it  may  be  considered 
to  be  due  to  small  forces  acting 
upon  every  element  of  mass  of 
the  body. 

Taking  the  latter  view,  let 
dF  be  the  resultant  force*  acting  upon  the  element  of  mass 
dm.  Then  since  dm  can  move  only  at  right  angles  t<>  the 
line  r,  which  joins  it  to  the  axis,  dF  must  be  perpendicular 
to  r.  When  the  body  is  given  an  angular  displacement  do, 
dm  is  displaced  through  ds,  therefore  the  work  done  by  dF  is 

d2W  =  dF  •  ds 
=  dF  -rdd 
=  dG  dd, 

*  dF  is  the  resultant  of  tin-  external  forces  which  ad  directly  on  • 
of  the  interna]  forces  which  an-  due  to  the  connection  of  dm  with  tin-  red  of 
the  body. 


Fig.  94. 


170  ANALYTICAL  MECHANICS 

where  dG  is  the  moment  of  dF  about  the  axis.  Thus  the 
total  work  done  by  all  the  forces  acting  upon  all  the  elements 
in  producing  the  angular  displacement  dd  is 

dW=  f  dG  dd 

-  do  f°fir       ^0  is  the  same  for  every 

Jo  element  of  mass.) 

=  Gdd, 

where  G  is  the  sum  of  the  moments  about  the  axis  of  all  the 
forces  acting  upon  the  elements  of  the  body,  i.e.,  the  result- 
ant torque  about  the  axis.  The  work  done  in  giving  the 
body  a  finite  angular  displacement  is,  therefore, 

W=  f°Gde  (III) 

Jo 

=  GO  [when  G  is  constant]. 

Therefore  work  done  by  a  constant  torque  in  producing  an 
angular  displacement  equals  the  product  of  the  torque  by  the 
angle. 

PROBLEMS. 

1.  A  weight  of  10  tons  is  to  be  raised  by  a  jackscrew.  The  pitch  of 
the  screw  is  \  inch  and  the  length  of  the  bar  which  is  used  to  turn  the  nut 
on  1  he  screw  is  2  feet  long.  Supposing  the  work  done  by  the  torque  to  be 
expended  entirely  against  gravitational  forces,  find  the  force  which  must 
he  applied  at  the  end  of  the  liar. 

2.  A  ball,  which  is  suspended  by  a  string  of  negligible  mass,  is  pulled 
aside  until  the  string  makes  an  angle  B  with  a  vertical  line.  Show  that 
the  work  done  is  the  same  whether  it  is  supposed  to  have  been  done  in 
rai.-iny;  the  hall  against  the  action  of  gravitational  forces,  or  in  rotating 
the  hall  and  the  string,  as  a  whole,  about  a  horizontal  axis  through  the 
point  of  suspension,  against  the  action  of  the  torque. 

3.  In  the  preceding  problem  take  the  following  data  and  calculate,  by 
both  methods,  the  amount,  of  work  done.     Weight  of  hall  =  12  ounces, 

length  of  string  -  :\  feet,  and  e  =  00°. 

4.  The  torque  which  has  to  he  applied  to  the  ends  of  a  rod  varies 
directly  with  the  angle  through  which  it  is  twisted;  derive  an  expression 
for  the  work  done  in  turning  one  end  of  the  rod  with  respect  to  the  other 
end  through  an  angle  6. 


WORK  171 

5.  In  the  preceding  problem  suppose  one  end  of  the  rod  to  be  fixed, 

while  the  other  end  is  firmly  attached  to  the  middle  of  another  rod  perpen- 
dicular to  it.  A  torqneof  10  pounds-foot  is  necessary  in  order  to  keep  the 
second  rod  in  a  position  turned  through  l">  about  the  axis  of  the  first 
rod.  How  much  work  must  be  done  in  order  to  produce  an  angular 
deflection  of  45°  ? 

6.  If  in  problem  5  the  torque  is  due  to  a  couple  the  forces  of  which  are 
applied  at  points  4  inches  from  the  axis  of  rotation,  find  the  forces  applied 
and  show  that  the  work  done  by  the  forces  equals  the  work  done  by  the 
torque. 

7.  Making  the  following  assumption  with  regard  to  the  normal  pressure 
at  the  bearings,  obtain  an  expression  for  the  work  done  in  giving  a  fly- 
wheel an  angular  displacement. 

(Hint.  —  For  this  and  the  following  problems  consult  §§-57  and  58.) 

(a)  Normal  pressure  is  constant. 

(b)  Vertical  component  of  the  total  reaction  is  constant. 

(c)  Normal  pressure  is  a  sine  function  of  the  angular  position;  the 
latter  being  measured  from  the  horizontal  plane  through  the  axis  of  the 
shaft. 

(d)  Normal  pressure  varies  as  the  square  of  the  sine  of  the  angular  posi- 
tion. 

8.  Find  the  work  done  in  giving  a  flywheel  a  complete  rotation.  The 
following  data  are  given.  The  flywheel  weighs  5  tons,  the  diameter  of 
the  shaft  is  10  inches,  the  coefficient  of  friction  in  the  journal  bearings  is 
0.05,  and  the  normal  pressure  on  the  bearings  satisfies  one  of  the  follow- 
ing conditions: 

(a)  Normal  pressure  is  constant. 

(b)  Vertical  component  of  the  total  reaction  is  constant. 

(c)  Normal  pressure  varies  as  the  sine  of  the  angular  position,  measured 
from  the  horizontal  plane  through  the  axis  of  the  shaft. 

(d)  Normal  pressure  varies  as  the  square  of  the  angular  position. 

9.  Derive  an  expression  for  the  work  done  in  giving  an  angular  dis- 
placement to  a  load  which  is  supported  by  a  fiat-end  pivot. 

10.  The  rotating  parts  of  a  water  turbine  which  weigh  50  tons  are  sup- 
ported by  a  flat-end  pivot.  The  diameter  of  the  shaft  is  10  inches  and 
the  coefficient  of  friction  is  0.().'5.     Find  the  work  lost  per  revolution. 

11.  Supposing  the  normal  pressure  to  be  constant  derive  an  expression 
for  the  work  done  in  giving  a  loaded  spherical  pivot  an  angular  displace- 
ment about  its  axis. 

12.  Supposing  the  normal  pressure  to  be  constant  derive  an  expi 

for  the  work  done  in  giving  a  loaded  conical  pivot  an  angular  displacement. 


172  ANALYTICAL  MECHANICS 

13.  A  vertical  shaft  carries  a  load  of  10  tons.  Find  the  work  lost  per 
revolution  if  the  ahafl  is  8  inches  in  diameter  and  has  a  flat-end  bearing; 
the  coefficient  of  friction  being  0.01. 

14.  Derive  an  expression  for  the  work  done  in  giving  a  collar-bearing 
pivot  an  angular  displacement. 

15.  A  vertical  shaft  carries  a  load  of  5  tons.  Find  the  work  lost  per 
revolution  if  the  shaft  is  supported  by  a  collar-bearing  pivot  which  has 
an  inner  diameter  of  0  inches  and  an  outer  diameter  of  8  inches.  The 
coefficient  of  friction  is  0.1. 

HOOKE'S  LAW. 

133.  Stress.  —  When  a  body  is  acted  upon  by  external 
forces  which  tend  to  change  its  shape  and  thus  give  rise  to 
forces  between  its  contiguous  elements,  the  body  is  said  to 
be  under  stress.     The  measure  of  stress  is  force  per  unit  area: 

S-j,  (IV) 

where  S  denotes  the  stress,  F  the  force,  and  A  the  area  on 
which  the  latter  acts. 

134.  Pressure,  Tension,  and  Shear.  —  Stresses  which  occur 
in  bodies  are  often  of  a  complex  nature,  but  they  may  be 
resolved  into  three  component  stresses  of  simple  character. 
These  are  called  pressure,  tension,  and  shear.  Pressure  tends 
to  compress,  tension  to  extend,  and  shear  to  distort  bodies. 
Shearing  stress  is  the  result  of  a  compressive  stress  com- 
bined with  a  tensile  stress  at  right  angles.  A  special  case 
of  -hear,  which  conies  into  play  within  a  shaft  when  the 
Latter  is  twisted,  is  called  torsion. 

135.  Strain.  -Strain  is  the  deformation  produced  by 
Btress.  The  measure  of  strain  is  the  percentage  deformation. 
For  instance,  if  the  deformation  consists  of  a  change  in  length 
the  Btrairj  equals  the  ratio  of  the  increase  in  length,  to  the 
original  length :  / 

f-j.  (V) 

win-re  s  denotes  the  stress,  /  the  increase  in  length,  and  L 
the  original  Length. 


WORK  1  ::* 

136.  Hooke's  Law.  —  The  relation  which  connects  a  stress 
with  the  strain  which  it  produces  is  known  as  Hooke's  law. 
It  states  that  stress  is  proportional  to  strain: 

S  =  \s,  (VI) 

where  X  is  the  constant  of  proportionality,  and  is  called  the 
modulus  of  elasticity. 

137.  Elastic  Limit.  —  Hooke's  law  holds  true  so  long  as 
stress  is  small  enough  to  leave  no  appreciable  permanent  def- 
ormation. In  other  words,  Hooke's  law  holds  true  strictly 
only  while  the  body  under  consideration  behaves  like  a  per- 
fectly elastic  body  under  the  action  of  the  given  stresses. 
All  bodies  are  more  or  less  imperfectly  elastic ;  that  is,  stresses 
always  leave  bodies  with  permanent  strains.  Therefore  at 
the  best  Hooke's  law  is  approximately  true  when  applied 
to  material  bodies.  The  approximation,  however,  is  close 
enough  for  practical  purposes  so  long  as  the  permanent  def- 
ormation is  negligible  compared  with  the  total  deformation 
produced  by  the  stress.  If  a  considerable  portion  of  the 
deformation  becomes  permanent  the  body  under  stress  is 
said  to  have  reached  its  elastic  limit,  when  Hooke's  law  does 
not  give  a  close  enough  approximation  and  consequent ly 
cannot  be  used. 

138.  Young's  Modulus.  —  The  modulus  of  elasticity  of  a 
body  which  is  being  stretched  is  called  Young's  modulus. 
Let  the  body  be  an  elastic  string,  a  wire,  or  a  rod,  and  let  .1 
be  the  area  of  its  cross-section,  L  its  natural  or  unstretched 
length,  and  I  the  increase  in  length  due  to  stretching.  Then 
we  have 

S  =  —  and  s  =  -  • 
.1  L 

Therefore  -  =Xt' 

A         I. 

,      F    L 

X-IT 


174 


ANALYTICAL  Mi;(  IIA.XK  s 


Thus  Young's  modulus  of  a  substance  equals,  numerically, 
the  force  necessary  to  stretch  a  uniform  rod  of  unit  cross- 
section,  which  is  made  of  the  given  substance,  to  double  its 
length.  During  the  process  of  stretching  Hooke's  law  is,  of 
course,  supposed  to  hold. 

139.  Work  Done  in  Stretching  an  Elastic  String.  —  Let  L 
denote  the  natural  length  of  the  string  and  x  its  length  at 
any  instant  of  the  process  of  stretching.  Then  the  work 
done  in  increasing  the  length  by  dx  is 

dW  =  Tdx 

=  AS  dx, 

where  T  is  the  tensile  force,  S  the  tension,  and  A  the  area  of 
the  cross-section  of  the  string.     But  by  Hooke's  law, 


Xs. 


In  this  case  s  = 


d\V  =  A.\' 


Therefore 
•-  L 


and 


W 


2Ll} 


-dx 


(x  —  L)  dx 


Vu..  95. 


where  x'  =  AX,  and  /  is  the  total  in- 
crease in  length.  Thus  the  work  done  varies  as  the  square 
of  the  increase  in  length.  Plotting  I  as  abscissa  and  IT  as 
ordinate  we  obtain  a  parabola,  Fig.  95. 

140.  Work  Done  in  Compressing  Fluids.  — Let  C,  Fig.  96, 
be  a  cylinder  which  contains  a  compressible  fluid  and  which 
is  provided  with  a  piston.  When  the  piston  is  displaced 
toward  the  left  work  is  done  against  the  force  with  which 
the  fluid  presses  upon  the  piston.     If  dx  denotes  the  dis- 


WORK 


175 


placement  and  F  the  force  on  the  piston  then  the  work 
done  is 


dW 


Fdx 

pA  dx 

pdv, 


where  p  is  the  pressure  in  the  fluid,  A  the  area  of  the  piston, 
and  dv  the  change  in  the  volume  of  the  fluid.  Therefore  t  he 
total  work  done  in  compressing  the  fluid  from  a  volume  i\ 
to  a  volume  r-2  is 

W=- J    pdv.  (1) 

When  the  law  connecting  p  and  v  is  given  the  work  done 
in  compressing  or  expanding  a  fluid  can  be  found  by  carry- 
ing out  the  integration 
indicated  in  equation  (1). 
During  expansion,  however, 
the  displacement  has  the 
same  direction  as  the  force 
which  causes  the  expansion ; 
therefore  the  sign  before  the 
integral  is  positive. 

141.  Representation  of  the 
Work  Done  in  the  PV-Dia- 
gram. —  When  the  volume 
of  the  expanding  fluid  is 
plotted  as  abscissa  and  the 
pressure  as  ordinate,  a  curve  is  obtained,  which  repre- 
sents, graphically,  the  law  connecting  j>  and  v.  Such  a 
representation  is  called  a  PV-diagram.  It  i-  evidenl  from 
equation  (1)  that  the  area  bounded  by  the  curve,  the  /-axis. 
and  the  two  vertical  lines  whose  equations  are  v  =  >\  and 
v  =  Vz,  represents  the  work  done  in  compressing  the  fluid 
from  Vi  to  Vi. 

142.  Isothermal  Compression  of  a  Gas. —  If  a  gas  is  com- 
pressed without  changing  its  temperature  the  compression 


Fig.  96. 


176  ANALYTICAL  -MECHANICS 

is  called  isothermal,  in  which  case  the  relation  between  p 
and  v  is  given  by  Boyle's  law,  i.e., 

pv-Jb.  (2) 

Substituting  in  equation  (1)  the  value  of  p  given  by  equation 
(2)  we  obtain 

w=-kT- 

=  Hog^-  (3) 

143.  Adiabatic  Compression  of  a  Gas. — If  no  exchange  of 
heat  is  allowed  between  the  gas  and  other  bodies  while  the 
former  is  being  compressed  the  compression  is  called  adiabatic. 
The  law  which  connects  p  and  v  in  an  adiabatic  compression 
or  expansion  of  a  gas  may  be  expressed  by  the  relation 

pvy  =  k,  (4) 

where  y  and  k  are  constants  for  a  given  gas.  Substituting 
in  ('(iiiation  (1)  the  value  of  p,  which  is  given  by  equation 
(4),  we  obtain 

W=-k  f* 

7-  1 
7-  1 

144.  Modulus  of  Elasticity  of  a  Gas.  —  Let  —  dv  denote  the 
change  in  volume  due  to  an  increase  in  the  pressure  of  a  gas 

by  an  amount  dp.    Then  the  stress  is  dp  and  the  strain  -  — . 

v 

Therefore  by  Hooke's  law 

,       .  —  dv 
dp  =  \ 

V 

or  \=-vp.  (6) 

dv 


WORK  177 

The  modulus  of  elasticity  X  is  not  a  definite  constant  for 
a  given  gas,  because  the  value  of  -^  depends  upon  the  tem- 
perature and  the  amount  of  heat  of  the  gas.     Therefore  the 

state  of  a  gas  for  which  ~  is  calculated  should  be  stated  in 
dv 

order  that  the  value  of  X  may  have  any  meaning  at  all. 
There  are  two  states  for  which  X  is  calculated,  namely,  the 
isothermal  and  the  adiabatic  states. 

145.  Isothermal  Elasticity.  —  When  the  compression  is  iso- 
thermal 

pv  =  k 

and  <*!>__  A 


dv  v2 


V-  (7) 


Therefore  the  isothermal  elasticity  of  a  gas  numerically 
equals  the  pressure. 

146.  Adiabatic  Elasticity.  —  When  the  gas  is  compressed 
adiabatically 

pvy  =  k 

and  ^  =  -  fair*-1 

dv 

=  -  ypv~l. 
•••      X=7P.  (8) 

147.  Torsional  Rigidity  of  a  Shaft.  —  Suppose  the  upper 
end  of  the  cylinder  of  Fig.  97  to  be  rotated  about  the  axis 
of  the  cylinder  through  an  angle  0,  while  the  lower  end  is 
fixed,  and  consider  the  stresses  and  the  strains  in  the  cylin- 
der. It  is  evident  that  the  strain  is  nil  at  the  axis  and  in- 
creases uniformly  with  the  distance  from  the  axis.  Further 
the  strain  is  nil  at  the  lower  base  and  increases  uniformly 
with  the  distance  from  it.  Since  Hooke's  law  holds  these 
statements  are  true  with  regard  to  the  stress  in  the  cylinder. 


178 


ANALYTICAL  MECHANICS 


Let  dF  denote  the  force  acting  on  the  area,  on  the  upper 
base,  of   a  ring  of  radius  r  and  width  dr,  then  the  stress 


equals 


dF 


But  if  6  is  the  angle  of  twist  at  the  upper 


2  vr  •  dr 

base  and  /  the  length  of  the  cylinder,  then  the  strain  equals 

y.     Therefore  by  Hooke's  law 


dF     =xrd 
2irrdr         I 

In  this  case  X  is  called  modulus  of  shearing 
elasticity  or,  simply,  shear  modulus.  Solv- 
ing the  preceding  equation  for  dF  we  get 

dF=~6r"-dr. 

Therefore  the  torque  acting  upon  the  area 
of  the  ring  is 


dG  =  r  >dF 

_  2_ttX 

I 

2ttX 

G=~r 


6rs  dr. 


Jrzdr 
o 


=  X 


21 


(9) 


Fig.  97. 


where  Q  is  the  total  torque  applied  at  the  upper  end  and  a 
the  radius  of  the  cylinder.  Thus  the  torque  necessary  to 
produce  a  given  angle  of  twist  varies  directly  as  the  fourth 
power  of  the  radius  and  inversely  as  the  length.  On  the 
other  hand  for  a  given  shaft  the  torque  varies  directly  with 
the  angle  of  twist . 
The  torsional  rigidity  of  the  shaft  is  defined  as  the  torque 
ny  to  produce  a  unit  angular  twist;  therefore 
G 


R  = 


=  X 


Tra4 

21 


(10) 


WORK  179 

It  will  be  observed  that  the  torsional  rigidity  of  a  Bolid 
shaft  varies  directly  as  the  fourth  power  of  the  radius  and 
inversely  as  the  length. 

148.  Work  Done  in  Twisting  a  Rod.  —  Work  done  by  a 
torque  is  obtained  by  substituting  the  expression  for  the 
torque  in  the  work  equation.     Thus 

W=  f°Gdd. 

Jo 

=  Xf[J)dd  [byeq.  (9).] 

=  W  (ii) 

where  k=  X — -  • 


PROBLEMS. 

1.  What  are  the  dimensions  of  stress,  strain,  and  modulus  of  elasticity? 

2.  A  steel  rod  of  £-inch  radius  is  found  to  stretch  0.004  inch  in  10 
inches  of  its  length  when  a  load  of  10,000  pounds  is  gradually  applied. 
Find  the  Young's  modulus  of  the  rod. 

3.  The  Young's  modulus  of  a  brass  wire  is  10.8  X  1011  ™? .     Find 

cm.2 

the  load  (in  pounds)  necessary  in  order  to  produce  an  elongation  of  0.5  mm. 
in  1  meter.     The  diameter  of  the  wire  is  1  millimeter. 

4.  The  modulus  of  shearing  elasticity  of  a  steel  shaft  is  11  X  106  pounds 
per  square  inch.  What  force  acting  at  the  end  of  a  lever  30  inches  long 
will  twist  asunder  the  shaft  if  it  is  0.5  inch  in  diameter? 

6.  A  brass  rod,  4  feet  long  and  1 .5  inches  in  diameter,  is  twiste*  1  through 
an  angle  of  9°  by  a  force  of  1500  pounds  acting  G  inches  from  the  axis  of 
the  rod.  If  on  removal  of  the  stress  the  bar  recovers  its  original  posil  ion, 
calculate  the  modulus  of  shearing  elasticity  of  the  rod. 

6.  Taking  the  data  of  the  preceding  problem  find  the  force  necessary  to 
give  an  angle  of  twist  of  2°  to  a  rod  15  inches  lout:,  0.5  inch  in  diameter. 

7.  An  elastic  string  of  natural  length  /  is  stretched  to  twice  its  length 
when  it  supports  a  weight  Ii'.  The  ends  of  the  string  are  connected  to 
two  points  at  the  same  level  and  a  distance  d  apart,  while  the  weighl  II' 
is  attached  to  the  middle  of  the  string.  Find  the  position  of  equilibrium 
of  the  weight. 

8.  A  spider  hangs  from  the  ceiling  by  a  thread  which  is  stretched  by 
the  weight  of  the  spider  to  twice  its  natural  length.     Show  that  the  work 


180  ANALYTICAL  MECHANICS 

done  by  the  spider  in  climbing  to  the  ceiling  equals  J  mgh,  where  m  is  the 
mass  df  the  spider  and  h  its  distance  from  the  ceiling. 

9.  The  outer  end  of  a  rial  spiral  spring  is  fixed,  while  the  inner  end  is 
attached  to  the  center  of  a  bar  20  cm.  long,  in  such  a  way  that  the  bar  is 
parallel  to  the  plane  of  the  spring.  Two  forces  of  500  dynes  each  applied 
at  the  ends  of  the  bar,  at  right  angles  to  the  bar  and  parallel  to  the  plane 

of  the  spring,  can  keep  the  bar  turned  through  an  angle  of  -    radians. 

What  torque  must  be  applied  in  order  to  keep  the  bar  in  position  after 
giving  it  three  turns? 

10.  In  the  preceding  problem  find  the  work  done  in  giving  the  bar  three 
turns.     What  portion  of  the  total  work  is  done  in  the  last  turn? 

11.  Prove  that  the  following  is  the  expression  for  the  torsional  rigidity 
of  a  hollow  shaft: 

(a*  -  fr4) 


I! 


■2  1 


where  6  is  the  inner  radius  of  the  shaft,  while  the  other  letters  represent 
the  same  magnitudes  as  in  §  147. 

12.  Derive  expressions  for  the  saving  of  material  and  loss  of  rigidity 
due  to  making  a  shaft  of  a  given  external  diameter  hollow. 

13.  Find  the  value  of  the  quotient  of  the  inner  to  the  outer  radius 
which  will  make  the  quotient  of  the  saving  of  material  to  the  loss  of  rigid- 
ity a  maximum. 

14.  The  weight  and  the  length  of  a  shaft  are  fixed;  find  the  ratio  of 
the  inner  to  the  outer  diameter  which  will  make  the  rigidity  of  the  shaft 
a  maximum. 

15.  The  torsional  moment  which  a  shaft  has  to  withstand  and  the 
Length  are  fixed;  find  the  ratio  of  the  inner  to  the  outer  diameter  which 
will  make  the  weight  of  the  shaft  a  minimum 

VIRTUAL  WORK. 

149.  Principle  of  Virtual  Work. — The  concept  of  work 
enables  us  t<>  formulate  a  principle,  called  the  principle  of 
virtual  work,  which  can  be  applied  to  equilibrium  problems 
t"  greal  advantage. 

In  on  lor  i,,  derive  this  principle  consider  a  particle  which 
i-  in  equilibrium.  Evidently  the  resultant  force  acting  upon 
the  particle  is  nil  and  remains  nil  so  long  as  the  particle  is  in 
the  equilibrium  position.     But  when  the  particle  is  given  a 


WORK  181 

small  displacement,  the  resultant  force  assumes  a  value  dif- 
ferent from  zero.  If  the  displacement  is  small  enough,  bo 
that  the  departure  from  equilibrium  position  and  conse- 
quently the  resultant  force  remains  small,  the  displacement 
is  called  a  virtual  displacement  and  the  work  done  by  the 
resultant  force  virtual  work.  We  will  call  virtual  force  the 
small  resultant  force,  which  is  called  into  play  by  the  virtual 
displacement. 

Let  Fi,  F2,  etc.,  be  the  forces  under  the  action  of  which 
the  particle  is  in  equilibrium.  When  the  particle  is  given  a 
virtual  displacement  ds,  these  forces  are  changed,  in  general, 
in  magnitude  and  direction  so  that  a  virtual  force  dF  acts 
upon  the  particle  during  the  displacement.  Then  the  virtual 
work  is 

dF  •ds  =  F1>  dsi  +  F2  •  ds2  +  •  •  •  ,  (VII) 

where  dsy,  ds2,  etc.,  are  the  displacements  of  the  particle 
along  the  forces  Fh  F2}  etc.,  due  to  the  virtual  displacement 
ds.  But  since  the  left-hand  member  of  the  last  equation 
is  an  infinitesimal  of  the  second  order  while  the  terms  of  the 
right-hand  member  are  infinitesimals  of  the  first  order  we 
can  neglect  the  left-hand  member  and  write 

Fi  dsj.  +  F2ds2  +  ■  •  •   +  =  0.  (VIII) 

Equation  (VIII)  states:  when  a  particle  which  is  in  equi- 
librium is  given  a  virtual  displacement  the  total  amount  of  work 
done  by  the  forces  acting  upon  the  particle  vanishes.  This  is 
the  principle  of  virtual  work. 

The  principle  of  virtual  work  is  applicable  not  only  to 
particles,  but  also  to  any  system  which  is  in  equilibrium. 
If  the  system  is  acted  upon  by  torques  as  well  as  forces, 
then  the  sum  of  the  work  done  by  the  virtual  torques  and 
the  virtual  force-  vanishes: 

Fxdsl  +  F2ds2  +  ■  ■  ■  +Gidei  +  Gidei+  •••-(>.       IX 


182 


ANALYTICAL  MECHANICS 


ILLUSTRATIVE  EXAMPLES. 

1.  Supposing  the  weights  in  Fig.  98  to  be  in  equilibrium  and  the  con- 
tacts to  be  smooth,  rind  the  relation  between  the  two  weights 

If  U'i  is  given  a  virtual  displace- 
ment towards  the  left  along  the  in- 
clined plane,  then  the  virtual  work  is 

-  T  ds  +  Wi  •  ds  sin  a  +  N  •  0  =  0, 
or  T  =  W\  sin  a. 

But  T  =  W2. 

Therefore         W2  =  Witana. 

2.  Two  uniform  rods  of  equal  weight  W  and  equal  length  a  are  jointed 
at  one  end  and  placed,  as  shown  in  Fig.  99,  in  a  vertical  plane  on  a  smooth 
horizontal  table.  A  string  of  length  I  joins  the  middle  points  of  the 
rods.     Find  the  tension  of  the  string. 

The  following  forces  act  upon  each  rod  —  the  weight  of  the  rod,  the 
pull  of  the  string,  the  reaction  at  the  joint,  and  the  reaction  of  the  table. 
Suppose  a  slighl  displacement  to  be  given  to  the  system  by  pressing  down- 
ward at  the  joint.  The  work  done  by  the  force  which  produced  the  dis- 
placement equals  the  sum  of  the  work  done  by  the  other  forces  which  act 
upon  the  rods  during  the  displace- 
ment. Bui  since  both  the  force  ap- 
plied and  the  displacement  produced 
are  very  small  their  product  is  negli- 
gible. Therefore  the  sum  of  the  work 
done  by  all  the  other  forces  is  zero. 

The  reactions  at  the  ends  of  the 
rod-  do  not  contribute  to  the  virtual 
work  because  each  of  the  reactions  is 

perpendicular  to  the  corresponding 

surface  of  contact  along  which  the  displacement  takes  place.  Therefore 
the  weights  and  the  tensile  force  of  the  string  contribute  all  the  virtual 
work.      If  '//  and  <lh  denote,  respectively,   the  increase  in  length  of  the 

Btring  and  the  distance  through  which  the  centers  of  mass  of  the  rods  are 
I  during  the  virtual  displacement  the  virtual  work  takes  the  form 


»( 


f  +  Wdfc) 


0. 


But  from  the  liirure  / 
and  <//<  =  —  a 
obtain 


a  Bin  0,  and  h  =  (i  cos  0.     Therefore  dl  =  a  cos  Odd 
Making  these  substitutions  and  simplifying  we 

T  =  'Jirtane1. 


WORK 


\s:\ 


3.   Find  the  mechanical  advantage  of  the  jack-screw. 

Let  /'  be  the  pitch  of  the  screw,  I  the  length  of  the  lever  arm,  F  the  force 
applied  and  P  the  force  derived.     Then 
since  at  any  instant  the  system  is  supposed 
to  be  in  equilibrium  the  virtual  work,  due     ^"" 
to  a  small  displacement,  must  vanish.   Let  / 
dd  denote  a  small  angular  displacement   l\ 
and  dh  the  corresponding  rise  of  the  screw. 
Then  if  G  denotes  the  torque  applied  the 
virtual  work  takes  the  form 
GdB-Pdh  =  0. 


But  G  =  F  •  I  and  dh 


Fide 


Ppdd 

2tt 


p.    Therefore 


5 


Fig.  100. 


Hence  the  mechanical  advantage,  which  is  the  quotient  of  the  force  de- 
rived to  the  force  applied,  is 

P  _2irl 

F        v   ' 


PROBLEMS. 

1.  By  the  application  of  the  principle  of  virtual  work  derive  the  ex- 
pression for  the  mechanical  advantage  of 

(a)  the  lever; 

(b)  the  wheel  and  axle; 

(c)  the  hydraulic  press; 

(d)  the  pulley  (a)  of  problem  13  on  page  21; 
(c)  the  pulley  (b)  of  problem  13  on  page  21; 

(f)  the  pulley  (c)  of  problem  13  on  page  21 ; 

(g)  the  pulley  (d)  of  problem  13  on  page  21. 

2.  Apply  the  principle  of  virtual  work  to 

(a)  illustrative  problem  1  on  page  17; 

(b)  problem  4  on  page  20; 

(c)  problem  6  on  page  16; 

(d)  problem  16  on  page  47. 

3.  Four  rods  of  equal  weighl  W  are  freely  jointed  so  as  to  form  n 
square.  The  system  is  suspended  vertically  from  one  of  the  joints.  A 
string  of  negligible  weighl  connects  two  of  the  joints  so  as  to  keep  the 
square  shape  of  the  system.     Find  the  ten-ion  of  the  string. 


184 


ANALYTICAL  ME< '  1 1 ANICS 


4.  An  clastic  band  of  weight  W  and  natural  length  I  is  slipped  over  a 
smooth  circular  cone  the  axis  of  which  is  vertical.  The  force  necessary 
to  sintch  the  st ring  to  double  its  natural  length  equals  X.  Find  the 
position  of  equilibrium  and  the  tension  of  the  string. 

5.  Find  the  mechanical  advantage  of  the  following  machines. 


fa 


CHAPTER   IX. 

ENERGY. 

150.  Results  of  Work.  —  Consider  the  work  done  by  the 
engine  of  a  train  in  pulling  it  upgrade.  The  work  done  may 
be  divided  into  three  parts: 

(a)  Work  done  against  frictional  forces. 

(b)  Work  done  against  gravitational  forces. 

(c)  Work  done  against  the  kinetic  reaction. 

The  result  of  work  done  against  frictional  forces  is  heat.. 
The  amount  of  heat  generated  is  proportional  to  the  amount 
of  work  done.  The  heat  may  be  utilized,  at  least  theoreti- 
cally, to  do  work.  Thus  a  part,  if  not  all,  of  the  original 
work  may  be  recovered. 

The  apparent  result  of  the  work  done  against  the  gravi- 
tational forces  is  the  elevation  of  the  train  to  a  higher  level. 
The  work  done  may  be  recovered  by  letting  the  train  come 
down  to  its  former  level  and  thereby  do  work.  Therefore 
the  work  done  against  gravitational  forces  may  be  considered 
to  be  stored  up. 

The  apparent  result  of  the  work  done  against  the  kinetic 
reaction  in  accelerating  the  train  is  an  increase  in  the  ve- 
locity of  the  train.  The  work  done  may  be  recovered  by 
letting  the  train  overcome  a  force,  which  tends  to  reduce 
the  velocity  of  the  train  to  its  original  value  Therefore  in 
this  case  also  the  work  done  may  be  said  to  have  been 
stored  up.  In  fact  in  all  three  cases  the  work  done  is 
stored  up.  In  the  first  case,  however,  work  is  qoI  available 
as  readily  as  in  the  other  two  cases.     In  order  to  converl 

185 


1st)  ANALYTICAL   MECHANICS 

boat  into  work  special  moans,  such  as  heat  engines,  etc.r 
have  to  be  used,  which  do  not  belong  to  the  domain  of 
ordinary  mechanics;  therefore  work  done  against  frictional 
forces  is  considered  as  lost.  On  the  other  hand  work  which 
is  done  againsl  gravitational  forces  or  against  kinetic  reac- 
fcions  is  directly  available  for  mechanical  work. 

151.  Energy.  Potential,  Kinetic,  and  Heat  Energy. — Energy 
may  be  defined  as  work  which  is  stored  up.  Work  stored  up 
in  overcoming  kinetic  reactions  is  called  kinetic  energy.  Work 
stored  up  while  overcoming  nonfrictional  forces,  such  as 
gravitational  forces,  is  called  potential  energy.  Work  done 
while  overcoming  frictional  forces  is  called  heat  energy. 

152.  Transformation  of  Energy.  —  Potential,  kinetic,  and 
Tieat  energy  are  different  (at  least  apparently*)  forms  of  the 
same  physical  entity,  i.e.,  energy.  Energy  may  be  changed 
from  any  one  of  these  forms  into  any  other  form.  "Whenever 
such  a  change  takes  place  energy  is  said  to  be  transformed. 
Transformation  of  energy  is  always  accompanied  by  work. 
In  fact  the  process  of  doing  work  is  that  of  transformation 
of  energy.  The  amount  of  energy  transformed  equals  the 
amount  of  work  done. 

The  units  and  dimensions  of  energy  are  the  same  as  those 
of  work. 

KINETIC  ENERGY. 

153.  Kinetic  Energy  of  a  Particle.  —  By  definition  kinetic 
energy  equals  the  work  done  against  the  kinetic  reaction 
in  giving  the  particle  its  velocity.  Since  there  is  no  mo- 
tion along  the  normal  to  the  path  of  the  particle  no  work  is 
done  againsl  the  normal  component  of  the  kinetic  reaction. 
Therefore  we  need  only  consider  the  work  done  against  the 
tangential  component. 

■  ni  developments  in  physical  sciences  tend  to  show  thai  differences 
i  different  forms  of  energy  are  only  apparent  :uui  that  all  forms  of 
."■.  in  the  last  analysis,  kinetic. 


ENERGY  187 

Denoting  the  kinetic  energy  by  T  and  putting  the  defini- 
tion into  analytical  language  we  obtain 


=m£ 


ds  , 
dt 


(I) 


Therefore  the  kinetic  energy  of  a  particle  equals  one-half  the  prod- 
uct of  the  mass  by  the  square  of  its  velocity.  Since  both  m  and 
v2  are  positive,  kinetic  energy  must  be  a  positive  magnitude. 
The  kinetic  energy  of  a  system  of  particles,  therefore,  equals 
the  arithmetic  sums  of  the  kinetic  energies  of  the  individual 
particles.     Thus 

T  =  \  Zmv\  (II) 

"When  all  the  particles  of  the  system  have  the  same  velocity 
T=\Mv\ 

where  M  is  the  total  mass  of  the  system. 

154.  Work  Done  in  Increasing  the  Velocity  of  a  Particle.  —  If 
the  velocity  of  a  particle  is  increased  from  v0  to  v  then  the 
work  done  against  the  kinetic  reaction  equals  the  increase 
in  the  kinetic  energy  of  the  particle.  This  will  be  seen  from 
the  following  analysis : 

d\ 


f  Hv 


(h 


ml    vdv 


=  T-  To. 


(Ill) 


*  The  first  negative  sign  indicates  Mir  fad  thai  T  is  the  work  done  against 
ami  nol  by  the  kinetic  reaction-;. 

The  second  negative  sign  belongs  to  the  kinetic  reaction  as  it  was  explained 

in  Chapter  VI. 


188  ANALYTICAL  MECHANICS 


PROBLEMS. 

1.  Show  that  the  dimensions  of  work  and  kinetic  energy  are  the  same. 

2.  A  body  of  50  gin.  mass  starts  from  the  top  of  an  inclined  plane  10  m. 

lone,  and  arrives  al  the  bottom  with  a  velocity  of  300  — ;.    Find  the  aver- 
"'  sec. 

age  frictional  force.     The  angle  of  elevation  of  the  plane  is  30°. 

3.  A  body  of  100  gm.  mass,  which  is  projected  up  an  inclined  plane, 

arrives  a1   the  top  of  the  plane  with  a  velocity  of  150- — '-.     Find  the 

velocity  of  projection,  supposing  the  frictional  force  to  be  constant  and 
equal  to  10,000  dynes.  The  length  of  the  plane  is  5  inches,  and  the  angle 
of  elevation  is  30°. 

4.  A  bullet  enters  a  plank  with  a  velocity  of  1500  — -.  and  leaves  it 

sec. 

with  a  velocity  of  1350  — —  •     How  many  such  planks  can  the  bullet 

see. 

penetrate? 

6.  In  the  preceding  problem  find  the  average  resisting  force  which  the 
planks  offer.    The  bullet  weighs  \  ounce. 

6.  A  catapult,  which  consists  of  an  elastic  string  15  cm.  long,. with  its 
end-  tied  tu  the  prongs  of  a  forked  piece  of  wood,  is  used  to  throw  stones. 
What  velocity  will  it  give  to  a  stone  of  5  gm.  mass  when  stretched  to  twice 
its  natural  length.     The  modulus  of  elasticity  of  the  string  is  2  pounds. 

7.  The  kinetic  energy  acquired  by  a  weight  of  750  pounds  in  falling 
through  a  distance  of  4  feet  is  to  be  absorbed  by  a  helical  spring,  5  inches 
long.  Find  the  modulus  of  elasticity  of  the  spring  so  that  it  will  not  be 
compressed  more  than  1  inch. 

8.  Having  a  given  size  and  shape,  how  will  the  penetrative  power  of  a 
bullet  depend  (a)  on  its  weight,  and  (b)  on  its  velocity.  The  resisting 
force  is  supposed  to  be  constant. 

155.  Kinetic  Energy  of  a  Rigid  Body  Rotating  About  a  Fixed 
Axis.  Suppose  the  rigid  body  A,  Fig.  lOl^to  rotate  about 
an  axis  through  the  point  O,  at  right  angles  to  the  plane  of 
the  paper.  Consider  the  kinetic  energy  of  an  element  of 
i! Li—  dm  al  a  distance  r  from  the  axis.  If  v  denotes  the 
velocity  of  the  element  and  dT  its  kinetic  energy,  then 


=  \  r2u2dm,     [v=  rw] 


ENERGY 


is'.) 


where  w  is  the  angular  velocity  of  the  body, 
total  kinetic  energy  of  the  rotating  body  is 


Therefore  the 


o 


/•'-'  dm  •  ws 


(IV) 


where  /  is  the  rnoment  of  inertia 
of  the  body  about  the  axis  of 
rotation. 

Comparing  the  expression  for 
the  kinetic  energy  of  rotation 
with  the  expression  for  the  kinetic 
energy  of  translation. we  observe 
that  moment  of  inertia  plays  the 
same  role  in  motion  of  rotation  as  mass,  the  linear  inertia, 
plays  in  motion  of  translation. 

The  expression  for  the  kinetic  energjr  of  a  rotating  body 
may  be  put  in  a  little  different  form  by  substituting  for  / 
its  value  in  terms  of  the  moment  of  inertia  about  a  parallel 
axis  through  the  center  of  mass.     Thus 


Fig.  101. 


=  \{Ic  +  ma-)u- 

=  %ma2oo2+  \  7,.ar 

=  \mv?  +  \Ieu?i  (V) 

where  ve  is  the  velocity  of  the  center  of  mass.  We  have 
thus  divided  the  kinetic  energy  into  two  parts  —  (a)  kinetic 
energy  due  to  the  motion  of  translation  of  the  body  as  a 
whole  with  the  velocity  of  the  center  of  mass,  (b)  kinetic 
energy  due  to  the  rotation  of  the  body  about  an  axis  through 
the  center  of  mass. 

156.  Work  Done  in  Increasing  the  Angular  Velocity  of  a  Rigid 
Body.  —  It  was  shown  in  §  154  that  the  work  dour  againsl 
the  kinetic  reaction  of  a  particle  equals  the  increase  in  the 
kinetic  energy  of  the  particle.  Therefore  the  wrork  done 
against  the  kinetic  reaction  of  any  number  of  particles  is  the 


190  ANALYTICAL  MECHANICS 

sum  of  the  increments  in  the  kinetic  energies  of  the  individ- 
ual particles.     Therefore 

W=Z(±mv2-%mv02). 

When  the  particles  form  a  continuous  system,  we  can  replace 
the  particles  by  elements  of  mass  and  the  summation  sign  by 
the  integration  sign.     Thus 

W=  I    (3  v2  dm  -  |  v02  dm) 
=  h  \    (r2<°2  dm  —  r2o)02  dm) 

=  \  co2  /    r2dm-\^2  I    r2  dm 
Jo  Jo 

=   I  /or  -  \  /a>o2,  (VI) 


where  co0  and  w  are  the  initial  and  the  final  values  of  the 
angular  velocity  of  the  body.  Therefore  in  this  case  also 
work  done  equals  the  increase  in  the  kinetic  energy. 


PROBLEMS. 

1.  The  flywheel  of  a  metal  punch  is  4  feet  in  outside  diameter  and 
weighs  500  pounds.  What  must  be  its  initial  velocity  in  order  that  the 
punch  may  exert  a  force  of  25  tons,  through  a  distance  of  1  inch,  without 
reducing  the  speed  of  the  flywheel  by  more  than  25  per  cent?  Neglect 
the  effect  of  the  shaft,  and  consider  the  flywheel  to  be  a  disk. 

2.  The  power  of  a  15-ton  car  was  shut  off  and  the  brakes  were  put  on 
at  :i  time  when  the  car  was  making  50  miles  an  hour.  On  each  of  the  8 
wheels  a  normal  brake-shoe  force  of  5000  pounds  was  applied.  Find  the 
distance  covered  by  the  car  before  coming  to  rest.  The  diameters  of  the 
wheels  are  30  inches,  the  tracks  are  horizontal,  and  the  coefficient  of  fric- 
tion for  Hie  contacl  between  the  shoes  and  the  wheels  is  0.2. 

3.  A  L00-tOD  Locomotive  making  a  mile  a  minute  is  to  be  stopped  within 
5<mi  yards.  What  brake-shoe  force  must  be  applied?  The  diameters  of 
the  wheels  are  (i  feet.     The  coefficient  of  friction  is  0.3. 

4.  Find  the  amount  of  heat  which  would  be  generated  if  the  rotation 
of  the  earth  about  its  axis  were  stopped.     The  mean  density  of  the  earth 

-  5.5  *~,  the  radius  =  4000  miles;  1  calorie  =  4.2  (10)7  ergs. 


ENERGY  L91 

6.  How  many  cubic  miles  of  ice  could  be  melted  by  the  heat  computed 
in  the  preceding  problem?    The  latent  heal  of  ice  is  80  calorics  per  gram. 

6.  The  winder  of  a  spinning  top  is  a  helical  spring,  which  is  set  in  a 

Cylindrical  piece  1  inch  in  diameter.  When  the  winder  is  honked  to  the 
top  and  twisted  through  tc  radians  a  force  of  1  pound  has  to  be  applied  to 
the  cylinder  tangentially  in  order  to  keep  it  from  untwisting  itself.  After 
the  sprint;  is  given  a  twist  of  2\  turns  the  top  is  released.  Find  the  kinetic 
energy  the  top  would  acquire  if  there  wore  no  frictional  forces. 

7.  In  the  preceding  problem  find  the  angular  velocity  of  the  top  suppos- 
ing it  to  consist  of  a  circular  plate  of  2  inch  radius,  and  of  \  pound  weight. 

8.  If  the  top  of  the  preceding  problem  turns  for  2  minutes  before  stop- 
ping, find  the  mean  torque  due  to  friction  and  resistance;  also  find  the 
total  number  of  revolutions  made. 

9.  A  top  is  given  a  motion  of  rotation  by  pulling  at  a  string  wound 
around  it.  Derive  an  expression  for  the  energy  communicated,  (a)  when 
the  force  applied  to  the  string  is  constant;  (b)  when  it  varies  directly  with 
the  length  of  the  string  which  is  unwound. 


POWER. 

157.  Power.  —  Power  is  the  rate  at  which  work  is  done. 
When  put  into  the  language  of  calculus  this  definition  be- 
comes 

P  =  ^.  (VII) 

Power  is  a  scalar  quantity  and  has  the  dimensions  [ML-T~*\. 
The  C.G.S.  unit  of  power  is  the  erg  per  second.  This  unit 
is  too  small  for  engineering  purposes;  therefore  two  larger 
units  are  adopted,  which  are  called  the  watt  and  the  kilowatt. 
The  following  relations  define  these  units: 


1  watt  = 


1  joule 

1  sec. 

=  10'  SB. 

sec. 

1  kilowatt  =  10^  watts 
=  10'" ''r^- 


192  ANALYTICAL  MECHANICS 

The  British  unit  of  power  is  the  horse  power,  denned  by  the 
following  equation: 

1  H.P.  =  33,000ft" 


PROBLEMS. 

1.  Show  that  1  horse  power  equals  about  746  watts. 

2.  The  engine  of  a  train,  which  weighs  150  tons,  is  of  200  horse  power. 
Find  the  maximum  speed  the  train  can  attain  on  a  level  track  if  there  is  a 
constant  resisting  force  of  15  pounds  per  ton. 

3.  The  diameter  of  the  cylinder  of  a  steam  engine  is  9  inches,  and  its 
length  10  inches;  the  mean  effective  pressure  per  square  inch  is  90  pounds, 
and  the  number  of  revolutions  per  minute  is  100.  Find  the  indicated 
horse  power. 

4.  Each  of  the  2  cylinders  of  a  locomotive  is  16  inches  in  diameter, 
the  length  of  the  crank  is  9  inches,  the  diameter  of  the  driving  wheels  is 
ti  feet,  the  velocity  of  the  train  is  40  miles  per  hour,  and  the  mean  effective 
pressure  is  75  pounds  per  square  inch.     Find  the  power  developed. 

6.  A  train  weighing  125  tons  moves  at  the  rate  of  50  miles  an  hour, 
along  a  horizontal  road.  Find  the  power,  in  kilowatts,  transformed  by 
the  motors  of  the  electric  engine  which  pulls  the  train.  The  resistance 
is  10  pounds  per  ton. 

6.  Find  the  horse  power  developed  by  an  engine  which  moves  a  train  at 
the  rate  of  30  miles  an  hour  up  an  incline  of  1  in  300.  The  train  weighs 
120  tons  and  there  is  a  resistance  of  15  pounds  per  ton. 

7.  A  belt  traveling  at  the  rate  of  45  feet  per  second  transmits  100 
horse  power.  What  is  the  difference  in  tension  of  the  tight  and  the  slack 
Bides  <>f  the  belt.    The  width  of  the  belt  is  20  inches. 

8.  A  L50-horse-power  steam  engine  has  a  piston  18  inches  in  diameter 
which  makes  100  strokes  per  minute.  Find  the  mean  effective  pressure 
of  tin'  steam  in  the  cylinder.     The  length  of  the  stroke  is  24  inches. 

9.  The  average  flow  over  the  Niagara  Falls  is  10,000  cubic  meters  per 
Becond.  The  average  height  is  L60  feet.  Find  the  power,  in  kilowatts, 
which  could  be  generated  if  all  the  energy  were  utilized. 

10.  A  lire  engine  pumps  water  with  a  velocity  of  125  —  through  a 

sec. 
iiu/./.le  1  inch  in  diameter.      Find  the  horse  power  of  the  engine  required 
t<»  drive  the  pump,  if  the  efficiency  of  the  pump  is  75  per  cent  and  the 


EXEIU'.Y  193 

nozzle  is  15  feel  above  the  surface  of  the  reservoir  which  supplies  the 
water. 

11.  Find  the  power  of  a  machine  gun  which  projects  600  bullets  per 

minute  with  a  muzzle  velocity  of  500  — -  and  angular  velocity  of  600  ir  ra- 

sec. 

dians  per  second.     The  bullets  are  cylinders  0.9  cm.  in  diameter  and  15  gm. 

mass. 

12.  A  shaft  transmits  50  horse  power  and  makes  150  revolutions  per 
minute.     Express  the  torque  transmitted  in  pounds-foot  and  dynes-cm. 

13.  An  electric  motor  develops  25  kilowatts  at  000  revolutions  per 
minute.  Find  the  torque  on  the  rotating  armature  due  to  the  field  mag- 
nets.    Neglect  friction. 

14.  Find  the  power  of  a  clock  which  has  a  maximum  run  of  S  days. 
The  weight  which  moves  the  works  has  a  mass  of  10  kg.  At  its  highest 
position  the  weight  is  15  inches  above  its  lowest  position. 

15.  A  twin-screw  steamer  has  engines  of  20,000  horse  power  and  when 
working  at  full  power  the  engines  make  75  revolutions  per  minute.  Find 
the  torque  transmitted  by  the  shaft  of  each  screw. 

16.  The  pitch  of  the  screw  propeller  of  a  ship  is  25  feet.  The  power 
transformed  by  the  propeller  is  15,000  horse  power,  when  the  ship  makes 
20  knots.  Assuming  that  there  is  a  slip  of  10  per  cent  at  the  propeller 
screw  and  that  the  efficiency  is  0.75,  find  the  torque  transmitted  by  the 
shaft,  also  the  thrust  on  the  bearings. 

17.  A  feed  pump  delivers  water  into  a  boiler  at  the  rate  of  20  lbs.  an 
hour.  If  the  pressure  in  the  boiler  is  150  lbs.  per  square  inch  above  the 
atmospheric  pressure,  find  the  effective  horse  power  of  the  pump. 

POTENTIAL  ENERGY. 

158.  Configuration.  —  The  arrangement  of  the  parts  of  a 
system  is  called  the  configuration  of  the  system.  The  system 
which  consists  of  this  book  and  the  earth,  for  instance,  is  in 
one  configuration  when  the  book  is  on  the  desk  and  in  an- 
other configuration  when  it  is  on  the  floor.  During  thel  rans- 
fer  of  the  book  from  the  floor  to  the  desk  the  system  passes, 
continuously,  through  infinite  number  of  configurations,  be- 
cause the  book  occupies  infinite  number  of  differenl  positions 
relative  to  the  cart  h. 

159.  Conservative  Forces. --If  the  work  done  in  bringing 
a  system  from  one  configuration  to  another  configuration  is 


194  ANALYTICAL  MECHANICS 

independent  of  the  manner  in  which  the  change  of  configura- 
tion takes  place,  the  forces  acting  upon  the  system  are  said  to 
be  conservative  forces.  Gravitational  forces  are  examples  of 
conservative  forces.  This  is  evident  from  the  result  of  §  129 
where  it  was  shown  that  the  work  done  against  gravitational 
forces  in  taking  a  body  from  one  point  to  another  is  inde- 
pendent of  the  path  along  which  the  body  is  carried. 

160.  Dissipative  Forces.  —  Forces  which  are  not  conserva- 
tive  are  called  dissipative  or  nonconservative  forces.  All  fric- 
tional  and  resisting  forces  are  of  this  type. 

161.  Potential  Energy.  —  The  potential  energy  of  a  system 
in  any  configuration  equals  the  work  done  against  the  con- 
servative forces  which  act  upon  the  system,  in  bringing  it 
from  a  standard  configuration  to  the  configuration  in  ques- 
tion. For  instance,  if  the  unstretched  state  of  an  elastic  string 
is  taken  to  be  its  standard  configuration,  then  the  potential 
energy  of  the  string  at  any  stretched  state  equals  the  work 
done  in  producing  the  extension.  The  potential  energy  of 
this  book  when  on  the  table  equals  the  work  done  in  raising 
it  from  the  floor  to  the  table,  provided  the  book  is  considered 
to  be  at  the  standard  configuration  when  it  is  on  the  floor. 

The  selection  of  the  standard  configuration  is  quite  arbi- 
trary and  i-  a  matter  of  convenience  only. 

It  is  evident  from  the  definition  of  potential  energy  that 
it-  value  is  zero  at  the  standard  configuration. 

Comparing  the  definitions  of  potential  energy  and  of  con- 
servative forces  we  Bee  that  the  potential  energy  at  any  given 
configuration  is  independent  of  the  manner  in  which  the 
system  is  brought  from  the  standard  configuration.  This  is 
equivalent  to  stating  that  the  potential  energy  of  a  system 
depends  upon  its  configuration.  But  coordinates  define  the 
configuration  of  a  system;  therefore  potential  energy  is  a 
function  of  the  coordinates. 

If  the  Bea  level  is  taken  as  the  standard  configuration, 
i.e..  tin-  position  of  zero  potential  energy,  then  the  potential 


ENERGY  L95 

energy  of  a  body,  due  to  gravitational  forces,  is  a  function 
of  the  vertical  height  of  the  body  above  the  sea  level;  in 
fact  it  equals  mgh,  where  mg  is  the  weighl  of  the  body  and 
A  its  height  above  the  sea  Level. 

162.  Difference  of  Potential  Energy.  —  The  difference  be- 
tween the  potential  energy  of  a  system  in  two  different  con- 
figurations equals  the  work  done  in  taking  the  system  from 
the  configuration  of  lower  potential  energy  to  that  of  higher 
potential  energy. 

Let  the  point  A,  Fig.  102,  represent  the  standard  con- 
figuration and  the  points  B  and  C  represent  two  other  con- 

,c 


Fig.  102. 

'A 

figurations.  Then  if  UB  and  Uc  denote  the  potential  energies 
at  B  and  C  respectively,  then  by  definition 

UB=WAB, 

Uc=WAC, 

where  IV Ab  and  WAc  are  equal,  respectively,  to  the  work 
done  in  going  from  A  to  B  and  from  A  to  C.     Therefore 

Uc-Ub=Wac-Wab=  Wbc,  (VIII) 

where  Wbc  equals  the  work  done  in  taking  the  Bystem  from 
B  to  C.  Thus  the  work  done  against  conservative  forces 
acting  upon  a  system  equals  the  increase  in  the  potential 
energy  of  the  system. 

163.  Isolated  System. — A  system  which  is  not  acted  upon 
by  external  force-  is  called  an  isolated  system.  An  isolated 
system  neither  gives  energy  to  external  bodies  nor  receives 
energy  from  them.  This  is  an  immediate  resull  of  the  defi- 
nition of  an  isolated  Bystem,  because  exchange  of  energj 


196  ANALYTICAL   MECHANICS 

presupposes  work  by  or  against  external  forces,  which  in  its 
turn  presupposes  interaction  with  external  bodies.  But 
since  no  external  forces  are  supposed  to  act  upon  the  system, 
there  cannot  be  interaction  with  external  bodies  or  exchange 
of  energy. 

164.  The  Principle  of  the  Conservation  of  Energy.  —  One  of 
the  greatest  achievements  of  the  nineteenth  century  was  the 
recognition  and  the  experimental  verification  of  the  great 
generalization  known  as  the  principle  of  the  conservation  of 
<  nergy,  which  states  that  the  total  amount  of  energy  of  an 
isolated  system  is  constant. 

By  means  of  the  interaction  of  the  different  parts  of  an 
isolated  system  the  various  forms  of  its  energy  may  be 
changed  into  other  forms,  and  the  distribution  of  the  energy 
within  the  system  may  be  altered,  but  the  total  amount 
of  energy  remains  constant.  In  other  words,  energy  may  be 
transformed  or  transferred  but  cannot  be  annihilated  or  created. 

165.  Dynamical  Energy.  —  Kinetic  and  potential  forms  of 
energy  are  called  dynamical  energy.  The  distinction  be- 
tween dynamical  and  nondynamical  energy,  such  as  heat 
energy,  chemical  energy,  etc.,  is  a  matter  of  convenience. 
Beal  energy  may  be  treated  as  kinetic  energy,  but  in  order 
to  do  that  molecules  and  their  individual  motions  have 
to  be  taken  into  account.  On  the  other  hand  chemical 
energy  may  be  treated  as  potential  energy  if  molecular  and 
atomic  forces  can  be  taken  into  account.  It  is  to  avoid  the 
complications  of  the  molecular  structure  of  bodies  that  these 
forms  <>l  energy  are  considered  as  nondynamical. 

166.  Conservation  of  Dynamical  Energy.  —  When  all  the 
forces  acting  within  an  isolated  system  are  conservative 
tin'  interchange  of  energy  is  confined  to  the  potential  and 
kinetic  forms  of  the  energy  of  t lie  system.  Therefore  apply- 
ing  the  general  principle  of  the  conservation  of  energy  we 
Bee  thai  in  Buch  a  system  the  sum  of  the  dynamical  energy 
remains  constant,  that  is, 

T+U=  const.  (IX) 


ENERGY  197 

If  T0  and  U0  denote  the  initial  values  of  T  and  U,  thou  the 
last  relation  gives 

T+U=  To+l'o 
and  T-T0=-(U  -U0).  (X) 

Therefore  if  only  conservative  forces  act  between  the  various 
parts  of  an  isolated  system,  the  sum  of  the  potential  and 
kinetic  energies  of  the  system  remains  constant,  in  other 
words,  the  gain  in  the  kinetic  energy  equals  the  loss  in  the  poten- 
tial energy.  Equation  (X)  will  be  called  the  energy  equal  ion. 
167.  Conservation  of  Dynamical  Energy  and  the  Law  of  Action 
and  Reaction.  —  The  principle  of  the  conservation  of  dynam- 
ical energy  may  be  obtained  from  the  Law  of  Action  and 
Reaction.  In  order  to  prove  this  statement  consider  an 
isolated  conservative  system.  Suppose  the  configuration  of 
the  system  to  have  changed  under  the  action  of  its  inter- 
nal forces.  Let  L'0  and  U  be  the  potential  energies  in  the 
initial  and  final  configurations,  respectively.  Then  the 
change  in  the  potential  energy  is 
(U-U0). 

During  the  change  in  the  configuration  of  the  system  the 
positions  and  the  velocities  of  the  particles,  which  form  the 
system,  undergo  changes.  Therefore  let  s0  and  s  denote 
the  positions,  and  v0  and  v  the  velocities  of  any  particle  in 
the  initial  and  final  configurations  of  the  system.  Further 
let  F  denote  the  resultant  force  which  acts  upon  the  particle. 
Then  the  change  in  the  potential  energy  of  the  system  due 
to  the  displacement  of  the  particle  from  s0  to  s  is 

-  f'Frds* 


f 


where  FT  is  the  tangential  component  of  the  force.     The 
normal  component  contributes  nothing  to  the  work.     There- 

*  Potential  energy  is.  by  definition,  the  \\<>rk  done  by  external  forces  against 
internal  forces.  Therefore  when  the  change  in  potential  energy  is  obtained  l>y 
computing  the  work  dune  by  internal  forces  1 1 » *  -  result  is  the  negative  of  Mir 
change  in  the  potential  energy.     Bence  the  negative  sign. 


198  ANALYTICAL  MECHANICS 

fore  the  total  change  in  the  potential  energy  of  the  system 
equals  the  Bum  of  the  work  done,  during  the  rearrangement, 
on  all  the  particles  of  the  system;  i.e., 

(U-U0)  =  -2  fFTds, 

where  the  summation  covers  all  the  particles  of  the  system. 
Therefore  substituting  the  expression  for  Fr)  which  was 
obtained  by  applying  the  law  of  action  and  reaction  to  the 
motion  of  particles,  we  obtain 

=  —  I!  m  I    vdv 

=  —  H  {\mv~—  ^mv02) 

=  -p(imt;2)-2(imt>o2)] 

=  -(T-T0), 

where  7',  and  T  are,  respectively,  the  values  of  the  total 
kinetic  energy  of  the  system  in  the  initial  and  in  the  final 
configurations.     Rearranging  the  terms  of  the  last  equation 

we  gel 

U+  T=  r0+  T0=  const. 

which  is  the  principle  of  the  conservation  of  dynamical 
energy.  Therefore  the  principle  of  the  conservation  of 
dynamical  energy  and  the  law  of  force  are  not  independent 
of  each  other  but  form  two  different  aspects  of  the  same 
universal  principle. 

[LLUSTRATIVE   EXAMPLE. 

Taking  into  account  the  variation  of  the  gravitational  attraction  with 
the  distance  <>f  a  body  from  the  center  of  the  earth,  find  the  potential 
energy  <>f  a  body  with  reaped  to  the  surface  of  the  earth. 

de  the  earth  the  weight  of  a  body  varies  inversely  as  the  square 
distance  from  the  center  of  the  earth.    Therefore  denoting  this 
variable  weighl  by  F  we  have 

r-k. 


ENERGY 


I'.V.t 


where  k  is  a  constant  and  r  is  the  distance  of  the  body  from  the  centei  of 

the  earth.  But  at  the  surface  of  the  earth 
the  weight  of  the  body  is—  mg,  therefore  F 
=  —  mg  when  r  =  a,  where  a  is  the  radius  of 
the  earth.  Therefore  making  these  substitu- 
tions in  the  last  equation  we  obtain 

k 

mga2. 


—  mq  =  —,  or  k  =  — 
y      a2' 

Therefore 

r2 

and 

U  =  f'Fdr 

,  frdr 
=  -mga-Ja   - 

=  mgaia-l} 

Discussion.  —  Plotting  the  potential  energy  as  abscissa  and  the 
height  above  the  surface  of  the  earth  as  ordinate  we  obtain  the  curve  of 
Fig.  103,  where  the  circle  represents  the  earth. 

When  r  =  a,  U  =  0;  as  it  should.  When  r  =  oo  ,  U  =  mga.  Therefore 
mga  is  the  maximum  value  of  the  potential  energy.  In  the  figure  this  is 
evident  from  the  fact  that  the  curve  approaches  asymptotically  to  the 


line  U  =  mga.     Whenr=  2  a,  U 


mga 
2 


Therefore  at  a  height  of  about 


4000  miles  the  potential  energy  equals  half  its  maximum  value. 

It  will  be  seen  from  the  following  analysis  that  for  small  heights  the 
potential  energy  may  be  considered  to  increase  linearly  with  h,  where  h  is 
the  height  above  the  surface  of  the  earth: 


U  =  mgaia-l 
=  mgail-a-Th) 


a  +  h 
=  mgh,     when     h  <^a* 


*  The  symbols  "<^C"  and  ";§>"\vil!  be  used  to  denote  greal  inequalities. 
Thus  "h<g!a"  should  be  read  "h  is  negligible  compared  with  o"  or  "h  is 
very  small  compared  with  a."  <  >n  the  other  hand  "a  ;§>/<"  should  be  read 
"a  is  very  large  compared  with  h." 


JOO 


ANALYTICAL  MECHANICS 


PROBLKMS. 


1.  A  reservoir  which  is  50  feet  long,  40  feet  wide,  and  10  feet  deep  is 
full  of  water.  Find  the  potential  energy  of  the  water  relative  to  a  plane 
25  feet  below  the  bottom  of  the  reservoir. 

2.  A  particle  slides  down  a  curve  in  a  vertical  plane  and  "loops  the 
Loop."  Find  the  minimum  height  the  starting  point  can  have  above  the 
center  of  the  "loop."     The  radius  of  the  "loop"  is  15  feet. 

3.  Find  the  least  velocity  with  which  a  bullet  will  have  to  be  projected 
from  the  earth  so  that  it  will  never  return  again. 

4.  A  uniform  rod  which  is  free  to  rotate  about  a  horizontal  axis  is 
held  in  a  horizontal  position.  With  what  angular  velocity  will  it  pass  the 
vertical  position  if  it  is  let  go? 

6.  A  cylinder  of  mass  m  and  radius  a  is  rotating  about  a  horizontal 
a\i>,  making  //  turns  per  second.  How  high  can  it  raise  a  mass  m',  which 
is  suspended  from  the  cylinder  by  means  of  a  string  of  negligible  mass? 

6.  A  particle,  which  is  attached  to  a  point  by  a  string  of  negligible 
ma--,  has  just  enough  energy  to  make  complete  revolutions  in  a  vertical 
circle.     Find  the  velocity  at  the  highest  and  at  the  lowest  points. 

7.  In  the  preceding  problem  show  that  the  tension  of  the  string  is  zero 
when  the  particle  is  at  the  highest  point  and  six  times  the  weight  of  the 
particle  when  it  is  at  the  lowest  point. 

8.  A  particle  starts  from  rest  at  the  highest  point  of  a  smooth  sphere 
and  slides  down  under  its  own  weight.     Where  will  it  leave  the  sphere? 

9.  A  particle  which  is  suspended  by  means  of  a  string  is  pulled  to  one 
side  until  it  makes  an  angle  a  with  the  vertical,  and  then  it  is  let  go.  Find 
the  position  at  which  the  tension  of  the  string  equals  the  weight  of  the 
particle. 

10.  In  the  preceding  problem  show  that  the  total  energy  remains  con- 
Btanl  during  the  motion  of  the  particle.  Also  find  the  velocity  at  the 
Lowest  position  when  a  =  GO0. 

11.  Supposing  the  tensile  force  necessary  to  stretch  an  clastic  string 
to  be  proportional  to  the  increase  in  length,  derive  an  expression  for  tho 
potential  energy  of  a  stretched  string. 

12.  Derive  an  expression  for  the  potential  energy  of  a  watch  spring. 


ENERGY  201 

GENERAL  PROBLEMS. 

1.  What  should  be  the  tractive  force  of  a  locomotive  in  order  that  it 
may  be  able  to  give  a  train  of  L50  tons  a  velocity  of  45  miles  per  hour  within 
one  mile  from  the  start?  The  resistance  per  ton  is  given  in  pounds  by 
the  numerical  relation  R  =  5  +  0.4  v'2,  where  v  is  the  velocity  in  miles  per 
hour. 

2.  In  the  preceding  problem  find  the  limiting  velocity. 

3.  The  effective  horse  power  of  a  vertical  water  wheel  is  46  and  its 
efficiency  is  70  per  cent.  If  the  head  of  water  is  25  feet  find  the  number 
of  gallons  of  water  which  have  to  be  delivered  to  the  wheel  per  minute. 

4.  A  belt  running  at  a  speed  of  1500  feet  per  minute  transmits  25  horse 
power.  Assuming  the  tensile  force  on  the  tight  side  of  the  belt  to  be 
twice  that  on  the  slack  side,  find  both  tensile  forces. 

5.  In  the  preceding  problem  find  the  width  of  the  belt  if  the  safe 
tensile  force  is  75  pounds  per  inch  width  of  the  belt. 

6.  Find  the  power  which  may  be  transmitted  by  a  belt  under  the 
following  conditions: 

The  width  of  the  belt  is  10  inches. 

The  pulley  which  the  belt  drives  is  4  feet  in  diameter  and  makes 

125  revolutions  per  minute. 
The  arc  of  contact  subtends  an  angle  of  150°  at  the  center. 
The  coefficient  of  friction  is  0.4. 
The  tensile  force  of  the  belt  is  not  to  exceed  90  pounds  per  inch 

of  width. 

7.  In  the  preceding  problem  find  the  tensile  force  on  the  slack  side 
of  the  belt. 

8.  In  problem  6  suppose  the  arc  of  contact  to  subtend  120°  at  the 
center  of  the  pulley. 

9.  In  the  preceding  problem  find  the  tensile  force  per  inch  width  of 
the  belt  on  the  slack  side. 

10.  Find  the  power  lost  due  to  friction  in  the  bearings  of  a  flywheel 
under  the  following  conditions: 

The  journals  are  6  inches  in  diameter  and  ID  inches  long. 

The  coefficient  of  friction  is  0.004. 

The  flywheel  weighs  15  tons  and  makes  200  revolutions  per 

minute. 
The  normal  pressure  on  the  bearings  is  constanl  over  the  surface 

of  contact. 

11.  In  the  preceding  problem  suppose  the  vertical  component  of  the 
total  reaction  to  be  constant. 


202  ANALYTICAL  MECHANICS 

12.  Find  the  power  lost  due  to  friction  in  the  bearings"  of  a  water 
turbine  under  the  following  conditions: 

The  rotating  system,  which  weighs  50  tons,  is  supported  by  a 

llat-eud  pivot  bearing  10  inches  in  diameter. 
The  coefficient  of  friction  is  0.01. 
The  turbine  makes  250  revolutions  per  minute. 

13.  In  the  preceding  problem  suppose  the  shaft  which  carries  the 
rotating  system  to  be  hollow,  with  an  inner  diameter  of  6  inches  and  outer 
diameter  of  12  inches. 

14.  In  problem  12  suppose  the  bearing  to  be  a  hemispherical  pivot 
with  constant  normal  pressure. 

16.  In  the  preceding  problem  suppose  the  vertical  component  of  the 
normal  pressure  to  be  constant. 


CHAPTER   X. 
FIELDS    OF   FORCE    AND    NEWTONIAN  POTENTIAL. 

168.  Fields  of  Force.  —  If  a  particle  experiences  a  force 
when  placed  at  any  point  of  a  region  the  region  is  called  afield 
<>/  force  The  gravitational  field  of  the  earth,,  the  electrical 
field  of  a  charged  body,  and  the  magnetic  field  of  a  magnet 
are  examples  of  fields  of  force. 

169.  Potential  Energy  and  Fields  of  Force.  —  The  potential 
energy  of  a  system  is  due  to  the  overlapping  of  the  fields 
of  force  of  its  parts.  For  instance,  the  earth  and  the  moon 
are  not  connected  by  anything  material,  yet  they  form  a 
system  which  has  potential  energy,  because  they  are  in  each 
other's  gravitational  field  of  force.  The  fact  that  a  stretched 
elastic  string  has  potential  energy  seems  to  contradict  this 
statement,  but  this  contradiction  is  only  apparent.  The 
potential  energy  of  the  stretched  string  is  also  due  to  the 
overlapping  of  the  fields  of  force  of  its  parts.  In  this  case, 
however,  molecules  form  the  parts  of  the  system. 

170.  The  Principle  of  the  Degradation  of  Potential  Energy. 
—  Consider  a  body  which  is  displaced  under  the  action  of  the 
forces  of  a  field  of  force.  A  certain  amount  of  work  is  done 
during  the  displacement.  If  the  body  is  not  acted  upon  by 
forces  which  are  external  to  the  field,  then  by  the  principle 
of  the  conservation  of  energy,  the  energy  of  the  body  remains 
constant  during  the  displacement.  Therefore  the  amount 
lost  by  one  form  of  the  energy  of  the  body  is  gained  by  the 
other.  The  work  done  is  the  measure  of  the  amount  of  the 
energy  transformed. 

The  principle  of  the  conserval  ion  of  energy  docs  nut  throw 
any  light  on  the  question,  "Which  form  of  energy  is  the 

203 


204  ANALYTICAL  MECHANICS 

loser  and  which  the  gainer?"  It  merely  states  that  the  loss 
equals  the  gain.  In  order  to  know  the  direction  of  the  trans- 
formation we  have  to  appeal  to  another  principle;  i.e.,  the 
principle  of  the  degradation  of  potential  energy,  which  states  : 

A  body  which  is  free  to  move  in  afield  of  force  moves  in  such 
a  may  as  to  diminish  its  potential  energy  * 

This  principle  is  nothing  more  or  less  than  a  simple  state- 
ment of  human  experience  with  things  that  "run  down." 
The  principle  states  that  water  flows  down  hill  under  the 
action  of  gravitational  forces,  that  a  clock  runs  down,  etc. 

171.  Force  Experienced  by  a  Particle  in  a  Field  of  Force.  — 
Consider  a  particle  in  a  field  of  force.  When  the  particle 
is  displaced  through  a  distance  ds,  under  the  action  of  the 
forces  of  the  field,  a  certain  amount  of  work  is  done  which 
equals  Fds,  where  F  is  the  resultant  force  due  to  the  field. 
Therefore,  by  the  principle  of  the  degradation  of  energy,  the 
potential  energy  of  the  particle  is  diminished  by  an  amount 
equal  to  Fds. 

Let  the  rate  of  increase  of  U  along  the  direction  of  the 

displacement  be  denoted  by  — ,  then-  —  ds  is  the  diminu- 
tion in  the  potential  energy.  Therefore,  equating  the  work 
jjoneby  the  forces  of  the  field  to  the  diminution  of  the  poten- 
tial energy  of  the  particle,  we  get 

Fds=  —  — ds, 
ds 

F=-^-t  (I) 

dS 

•  Thia  principle  may  be  called  the  dynamical  version  of  the  second  law 
of  thermodynamics. 

I-    bould  l»-  remembered  thai  the  forces  which  enter  into  the  equations 

|  odF      -  .  ■  are  equal  but  oppositely  directed.     In  the  second 

equation  /•'  represents  the  resultanl  force  which  a  particle  experiences  by  virtue 

potential  energy.    <  >n  the  other  hand  Pin  the  definition  of  potential 

lenotee  the  external  force  which  has  to  be  applied  to  the  particle  in 


FIELDS  OF  FORCE  AND   NEWTONIAN  POTENTIAL     205 

Splitting  equation  (I)  into  three  component  equations  we 
have 

dx 


dz 


(I') 


Equations  (I)  and  (F)  state  that  2/ie  force  along  a  given  direc- 
tion which  a  particle  experiences  by  virtue  of  its  potential  energy 
equals  the  rate  of  diminution  of  the  potential  energy  along  the 
given  direction. 

172.  Torque  Experienced  by  a  Rigid  Body  in  a  Field  of  Force. 
—  Suppose  the  rigid  body  to  be  displaced  under  the  action  of 
the  forces  of  the  field  through  an  angle  dd.  Then  an  amount 
of  work  G  dd  is  done,  where  G  is  the  torque  which  the  body 
experiences  in  the  field.  By  the  principles  of  the  conserva- 
tion and  degradation  of  dynamical  energy  this  work  must 
come  from  the  potential  energy  of  the  body  in  the  field. 
Therefore  denoting  the  rate  of  increase  in  the  potential  energy 

of  the  body,  due  to  a  rotation  about  a  given  axis,  by  - —  we 

dd 

have 

O-f.  (ID 

Equation  (II)  states  that  the  torque  which  a  rigid  body  exjn  ri- 
ences  by  virtue  of  its  potential  energy  equals  the  rate  of  diminu- 
tion of  this  energy  as  the  body  (urns  about  the  axis  of  the 
torque. 

on  lor  to  overcome  the  forces  winch  the  particle  experiences  because  of  its  po- 
sition in  a  field  of  fonc,  and  thereby  to  bring  the  particle  from  the  standard 
configuration  to  the  one  in  which  it  has  potential  energy  U. 


206 


ANALYTICAL   MECHANICS 


ILLUSTRATIVE   EXAMPLES. 

1.    Find  the  font-  which  :i  particle  placed  upon  a  smooth  inclined  plane 
experiences  by  virtue  of  its  potential  energy.    Also  find  the  components 

of  the  force  along  I  lie  axes  of  :i  rectangu- 
lar Bystem,  in  which  the  r-axis  is  normal 
to  the  inclined  plane  and  the  .c-axis  is 
horizontal. 

Let  the  origin  of  the  axes,  Fig.  104,  be 
the  position  of  the  particle.  Then  if  h 
denotes  ita  heighl  from  the  base  of  the 
inclined  plane  the  potential  energy  is  mgh. 
Therefore  the  force  along  the  vertical  is 
given  by 

dV 

dh  dh 


(mgh) 


mg. 


Fig.  104. 


Thus  the  force  due  to  the  gravitational  field  is  downwards  and  equals  the 
weight  of  the  particle.  The  components  of  the  force  are  found  by  equa- 
tion <F).     Thus 


x  =  _  ^ 


mg 


0. 


Therefore  the  force  along  the  .c-axis  is  nil. 

dh 


dU 


mg 


dy 


mg  bw  a. 


Therefore  the  component  of  the  force  along  the  plane  is  downwards  and 
equals  »'</  sin  a. 

„         dV  dh 

Z  =  -   —  =-mg  — 

dz  dz 


mg  cos  a. 


Therefore  the  component  along  the  :-axis  tends  to  move  the  particle  nor- 
mally into  the  plane  and  has  a  magnitude  equal  to  mg  cos  a.  The  com- 
ponents along  the  .r-axis  and  the  c-axis  produce  no  motion  because  X 
equals  zero  and  Z  is  exactly  balanced  by  the  reaction  of  the  plane.  The 
foregoing  results  may  be  verified  by  finding  the  components  of  mg  by  the 
common  method,  i.e.,  by  taking  projections  of  mg  along  the  axes. 

2.    A  rigid  body  which  is  free  to  rotate  about  a  horizontal  axis  is  dis- 
through  an  angle  0.     Find  the  torque  due  to  the  gravitational 
held  of  the  earth. 


FIELDS  OF  FORCE  AND  NEWTONIAN    POTENTIAL     207 


Let  .1,  Fig.  10"),  be  the  body,  0  the  point  where  the  axis  pierces  the 
plane  of  the  paper,  ( '  the  center  of  mass,  and  D  its  distance  from  the  axis. 
Then  at  the  displaced  position  the  potential 
energy  is 


Therefore   the 

hotly  is 


U  =  mgh 

=  mgD  (1  —  cos0). 

torque   experienced   by   the 


Fig.  105. 


—  —mgDmi  6. 

This  result  may  be  easily  verified  by  consid- 
ering the  moments  of  the  forces  which  act 
upon  the  body.  The  forces  which  act  upon  the  body  are  the  reaction  of 
the  axis  and  the  weight  of  the  body.  The  moment  of  the  reaction  is 
nil;  therefore  the  resultant  moment  is  entirely  due  to  the  weight  and 
equals 

G=  —  mg-d  =  —  mgDs'md, 

which  is  the  result  obtained  by  the  other  method.  The  negative  sign  i? 
introduced  to  indicate  the  fact  that  the  rotation  is  clockwise. 

173.  New  Condition  of  Equilibrium.  —  Equations  (I),  (I'), 
and  (II)  provide  us  with  a  new  condition  for  the  equilibrium 
of  conservative  systems.  It  was  shown  in  Chapters  II  and 
III  that  a  system  is  in  equilibrium  when  the  resultant  force 
and  the  resultant  torque  vanish. 

Therefore  setting  F  and  G  equal  to  zero  in  equations 
(I)  and  (II)  we  obtain 

dU 

ds 

dU 

dd 

where  the  differentiation  in  the  first  equation  is  with  respect 
to  any  direction  and  that  in  the  second  with  respect  to  an 
angle  about  any  axis.  Hut  when  equations  (III)  are  satis- 
fied, U  has  a  stationary  value,  that  is,  the  value  of  £/is  either 
a  minimum,  or  a  maximum,  or  a  constant.     Therefore  the 


i). 


=  0, 


(III) 


208  ANALYTICAL  MECHANICS 

new  condition  states  —  in  order  that  a  conservative  system  be  in 
equilibrium  its  potential  energy  must  have  a  stationary  value. 

174.  Analytical  Criterion  of  Stability.  —  The  equilibrium  of  a 
body  is  said  to  be  stable  if  it  is  not  upset  when  the  body  is 
given  a  small  displacement. 

Potential   energy  is  a  function  of  coordinates,  therefore 
we  <aii  denote  the  potential  energy  of   a  particle   at   the 
point   (rri,  yh  Zi)  by  the  functional  relation 
Ui  =  U(xuyi,Zi). 

Let  us  suppose  the  point  (xh  yh  Zi)  to  be  a  position  of  equi- 
librium of  the  particle,  and  investigate  the  stability  of  the 
equilibrium.  If  the  particle  is  given  a  displacement  8x,  the 
potential  energy  in  the  new  position  becomes 

U2  =  U(x1+8x,yhz1). 

Expanding  U2  by  Maclaurin's  theorem  in  powers  of  8x  we 

obtain 


a-HSt"'©^ 


where  the  subscripts  after  the  parentheses  denote  that  after 
the  indicated  differentiations  are  carried  out  the  coordinates 
x,  y,  and  z  must  be  replaced  by  x\,  yi,  and  Z\,  which  are  the 
coordinates  of  the  equilibrium  position. 

But  since  the  particle  is  in   equilibrium   at   the   point 


dXh 


( 

Since  o.r,  the  displacement,  is  small  we  can  neglect  all  the 
terms  of  the  right-hand  member  of  the  last  equation  except 
the  first.    This  gives 


FIELDS  OF  FORCE  AND  NEWTONIAN   POTENTIAL     209 

Case  I.  —  Suppose  ( — -)  to  be  positive.     Then  U%  —  Uy  is 
\oX  l\ 

positive  and  consequently  Ux  is  a  minimum.  But  according 
to  the  principle  of  the  degradation  of  energy  a  body,  which  is 
free,  moves  in  such  a  way  as  to  diminish  its  potential  energy. 
Therefore  when  the  force  which  produced  the  displacement  5.r 
is  removed  the  particle  returns  to  the  point  (xu  yu  Zi),  where 
its  potential  energy  is  a  minimum.  Evidently  the  equilib- 
rium is  stable  in  this  case. 

Case  II. — Suppose  ( — -)    to  be  negative.     Then  U2-U1 

is  negative  and  consequently  L\  is  a  maximum.  Therefore 
if  the  particle  is  given  a  small  displacement  8x  and  then  left 
to  itself,  it  will  move  away  from  the  point  (x\,  yi,  Zi),  where 
its  potential  energy  is  a  maximum.  In  this  case  the  equi- 
librium is  unstable. 

Case  HI.  —  Suppose  f  — ^J  to  be  zero.     There  are  throe 

special  cases  to  be  considered: 

(a)  The  order  of  the  first  differential  coefficient  which 
does  not  vanish  is  odd. 

(b)  The  order  of  the  first  differential  coefficient  which 
does  not  vanish  is  even. 

(c)  All  of  the  differential  coefficients  vanish. 

It  is  evident  that  when  (c)  is  true  the  potential  energy  of 
the  particle  has  a  constant  value  and  does  not  change  with 
the  position  of  the  particle.  Therefore  when  the  particle 
is  left  to  itself  after  giving  it  a  small  displacement  it  will 
neither  return  to  its  original  position  nor  go  on  changing 
its  position.  The  potential  energy  is  the  same  and  the 
particle  is  in  equilibrium  at  all  points.  In  this  case  the 
equilibrium  is  said  to  be  neutral  or  indifferent. 

It  may  be  shown  that  when  (a)  holds  the  equilibrium  is 
stable.  On  the  other  hand  when  (b)  is  true  the  equilibrium 
is  stable  or  unstable  according  as  the  first  differentia]  coeffi- 
cient which  does  not  vanish  is  positive  or  negative. 


210 


ANALYTK  !AL  MECHANICS 


The  three  types  of  equilibrium  are  illustrated  by  the  three 
equilibrium  positions,  Fig.  100,  which  a  right  cone  can  as- 
sume on  a  horizontal  plane. 


Fig.  106. 

NEWTONIAN  POTENTIAL. 

175.  Newtonian  Law  of  Force.  —  The  law  of  force  between 
1  wo  interact  ing  particles  is  called  a  Newtonian  law  of  force  if 
the  particles  attract  or  repel  each  other  with  a  force  which 
acts  along  the  line  of  centers  of  the  particles,  and  which 
varies  directly  as  the  product  of  the  masses  of  the  particles 
and  inversely  as  the  square  of  the  distance  between  them. 
The  forces  between  two  material  particles,  between  two 
small  electrical  charges,  and  between  two  small  magnetic 
poles  obey  the  Newtonian  law  of  force.  The  following  are 
the  familiar  forms  in  which  the  law  is  written  for  material, 
electrical,  and  magnetic  particles,  respectively, 


F  = 


F=  - 


F=  - 


unit 


(IV) 


where  -, .  k,  and  n  are  constants.     When  the  interacting  par- 
ticles  are  in  free  space  the  numerical  value  of  the  constants 

I:  and  n  is  unity,  while 

7=6.7X10-3      Cm'3   2- 
gm.  sec.2 

176.  Newtonian  Field  of  Force.  —  When  the  forces  which 
act  in  a  region  obey  the  Newtonian  law  of  force  the  region 
is  called  a  X<  wtonian  field  of  force. 


FIELDS  OF   FORCE  AND   XKWTOXIAN"    POTKXTIAI.      I'll 

177.  Newtonian  Potential.  —  The  potential  energy  of  a  unit 
mass  placed  at  a  point  of  a  Newtonian  field  is  called  the 

potential  at  that  point.  The  standard  configuration  or  the 
position  of  zero  potential  is  taken  to  be  infinitely  far  from 
the  center  of  the  field.  But  the  potential  energy  of  a  body 
equals  the  work  done  in  bringing  the  body  from  the  position 
of  zero  potential  energy,  therefore  the  following  definition  is 
equivalent  to  the  one  just  given. 

The  potential  at  a  point  equals  the  work  done  in  bringing  a 
unit  mass  from  an  infinite  distance  to  that  point. 

178.  Potential  Due  to  a  Single  Particle.  —  Let  m  be  the  mass 
of  the  particle,  U  the  potential  energy  of  a  particle  of  mass 
?n'  placed  in  the  field  of  force  of  the  first  particle,  r  the  dis- 
tance between  the  two  particles,  and  V  the  potential  at  the 
position  of  m'  due  to  m.     Then  by  the  definitions  of  V  and  U 


m 


=  h  fr(-F)dr 
where  F  is  the  force  experienced  by  m'  due  to  the  field  of  m 


(V) 


■D      4.  V  mm 

But  F  =  -7  — — 

r- 


Therefore  V=  ym  I    — 

t/oo     T 

—  T  =  .  (VI) 

The  negative  sign  indicate-  the  facl  that  when  a  particle  is 
brought  to  the  field  of  another  attracting  particle  work  will 
be  done  by  the  particle  and  not  by  the  agent  which  brings 
it.  Therefore  the  potential  due  to  a  material  particle,  as 
we  have  denned  it,  is  everywhere  negative,  excepl  at  infinity 
where  it  is  zero.  In  ease  of  electrical  and  magnetic  masses 
potential  is  defined  as  the  work  dime  in  bringing  a  unit  posi- 
tive charge,  or  unit  positive  pole,  from  infinity.     Therefon 


212  ANALYTICAL  MECHANICS 

the  potentials  due  to  a  negative  charge  and  a  negative  pole  are 
negative,  while  the  potentials  due  to  a  positive  charge  and  a 
positive  pole  arc  positive. 

179.  Potential  Due  to  Any  Distribution  of  Mass.  —  When  the 
field  of  force  or  the  potential  field  is  due  to  a  number  of  par- 
ticles (material,  electrical,  or  magnetic),  the  potential  at  a 
point  equals  the  algebraic  sum  of  the  potentials  due  to  the 
various  particles.  Thus  if  mx,  m2,  m3,  etc.,  be  the  masses  of 
the  particles  and  rly  r2,  n,  etc.,  their  distances  from  the  point 
considered,  then  the  potential  at  the  point  is 


V    n  r2 

„ni 


(VII) 


When  the  field  is  due  to  a  continuous  distribution  of  mass 
the  last  equation  may  be  put  in  the  form  of  an  integral. 
Thus 

dm 


rmdm 

i/0        T 


(VII') 


180.  Intensity  of  the  Field.  —  The  intensity  at  any  point  of 
a  potential  field,  or  a  field  of  force,  is  defined  as  the  force 
t.r/K  /•/'(  need  by  a  unit  mass  when  placed  at  that  point. 

Let  H  denote  the  intensity  at  a  point.  Then,  if  F  is  the 
force  experienced  by  a  mass  m'  when  placed  at  that  point, 
we  have,  by  definition, 

W  =  —,i  (VIII) 

m 

and  //  =  — , 

in 

=  --,(-) 

III    \dS  ) 

8»ta7 


FIELDS  OF  FORCE  AND  NEWTONIAN  POTENTIAL     213 


Similarly 


and 


= 

dV 

ds  ' 

Hx  = 

dV 
to' 

Hy  = 

dV 

ay' 

H.= 

dz  ' 

(IX) 


Therefore  the  component,  along  any  direction,  of  the  intensity 
at  any  point  equals  the  rate  at  which  the  potential  diminishes 
at  that  point  as  one  moves  along  the  given  direction. 


ILLUSTRATIVE  EXAMPLES. 


1.  Find  the  expressions  for  poten- 
tial and  intensity  at  a  point  due  to 
a  spherical  shell. 

Let  P,  Fig.  107,  be  the  point  and 
R  its  distance  from  the  center  of  the 
shell.  Then  taking  a  zone  for  the 
element  of  mass,  as  shown  in  the 
figure,  we  get 


Fig.  107. 


and 
Therefore 


dm  =  <r  •  2 xo  sin  6  •  ad9, 
r  =  V(R  -  ocos0)2  +  a2sin2£ 
-  Va2  +  R2-2aRcosd. 


simdild 


rm(lm 

'  — »/.  T 

fir  Sill    7  (10 

=  -7r27ra-Jo   Va»  +  i2*  -  2  a«  cos  0 

-  -  ^|^  [(a'  +  R*  -  2  aR  cos 9) l]0 

-  -  ^^[(a2  +  2 a/2  +  fi8)*  -  (a»-  2aR  +  «•)*].  ' 

There  are  two  different  cases  which  have  to  be  considered  separately. 


214 


ANALYTICAL  MECHANICS 


(a)  Point  Outside  the  Sphere.  —  In  this  case  R>a.    Therefore  the 
expression  for  the  potential  may  be  put  in  the  form 

v=_yrjJIa[{a  +  R)_iR_a)] 

T-iira2 
m 

Therefore  outside  the  shell  the  potential  is  the  same  as  if  the  mass  of  the 
shell  wnr  concentrated  at  its  center. 

(b)  Po»  i  Within  the  Sphere.  — In  this  case  R  <  a.     Therefore 
v  _      yT  2  ira  , 


R 

t  4  7ra2 


[(a  +  R)-(a-  R)\ 


-y 


H  =  - 


Therefore  within  the  shell  the  potential  is  constant  and  equals  that  at  the 
surface. 

If  //  denotes  the  intensity  of  the  field  due  to  the  shell,  then 

dv 

dR 

=  —  7  -=r-2   when  R  >a. 

=  0     when  R  <  a. 

Therefore  the  shell  attracts 
8  particle  which  is  outside 
with    the    same    foTCi 

all  of  its  mass  were  concen- 
trated at  its  center.    I  m  the 

other  hand  the  shell   exerts 
on  b  particle  which 

is  within  the  shell.    The  dis- 
tribution of  V  and  //  in  the 
field  are  represented  graph- 
ically  in    Fig.    Kis,   where 
the  po- 
ind |  II )  the  intensity. 
2.    I  m. I  the  expressions  for  the  potential  and  the  intensity  due  to  a 
i 


FIELDS  OF  FORCE  AND  NEWTONIAN  POTENTIAL     215 

There  are  two  cases  which  have  to  be  considered  separately. 
(a)  Point  Outside  the  Sphere.  —  Consider  the  sphere  to  be  made 
of  concentric  shells  of  thickness  dp.     Then,  since  the  point  is  outside  every 
one  of  these  shells  the  potential  due  to  any  one  of  the  shells  is,  according 
to  the  results  of  the  last  problem, 

...  dm 

d\   =  —  y—=ri 


where  dm  is  the  mass  of  the  shell  and  R  the  distance  of  the  point  from  the 
center.    Hence  the  potential  due  to  all  the  shells  in  the  sphere  is 

dm 


r-  am 


=  -yR' 

where  m  is  the  mass  of  the  sphere.  Therefore  the  potential  at  a  point 
outside  of  a  sphere  is  the  same  as  that  due  to  a  particle  of  equal  mass 
placed  at  the  center. 

(b)  Point  Within  the  Sphere.  —  In  this  case  we  divide  the  sphere 
into  two  parts  by  means  of  a  concentric  spherical  surface  which  passes 
through  the  point.  Then  the  potential  due  to  that  part  of  the  sphere 
which  is  within  the  spherical  surface  is  obtained  by  the  result  of  case  (a). 
Thus  if  nh  denotes  the  mass  of  this  part  of  the  sphere  and  V\  its  potential, 
then 

In  order  to  find  the  potential  due  to  the  rest  of  the  sphere  suppose  it 
to  be  divided  into  a  greal  number  of  concentric  spherical  shells.  Then 
since  every  one  of  the  shells  contains  the  point  the  potential  due  to  any 
one  of  them  is 

dVi  =  —  y  —  =  —  4  ivyrp  dp, 
P 

where  dm   is  the  mass,  p  the  radius,  and  dp  the  thickness  of  the  shell. 

Therefore  the  potential  due  to  all  the  shells  having  radii  between  R  and  a  is 


Vt  =  —  -iwyr  \    pdp 
=  —  2-rryT  (fl         R  :. 


21G 


ANALYTICAL   MECHANICS 


Therefore  the  potential  due  to  the  entire  Bphere  is 
V  =  Vi+  V, 


=  —  ym 


3  <r-  -  R- 

2  a3 


When  R  is  plotted  as  abscissa  and 
V  as  ordinate  the  distribution  of 
the  potential  is  given  by  a  curve 

similar  to  (I)  of  Fig.  L09. 

Now  consider  the  intensity  at  a 
point  in  the  field  of  the  sphere. 
(a)  Point  Outside  the  Sphere. 

H  =- 


dV 

dR 


Fig.  109. 


Therefore  the  distribution  of  the  field  intensity  outside  of  the  sphere  is 
tin-  -ame  as  that  due  to  a  particle  placed  at  the  center. 
(hi  Point  Within  the  Sphere. 

dY 

dR 


H  =-r= 


=  -y-R- 

<r 

Therefore  aithin  the  Bphere  the  distribution  of  the  field  intensity  obeys  the 

harmonic  law;  i.e.,  the  intensity  varies  directly  as  the  distance  from  the 
center.  In  Fig.  It)'.),  curve  (11)  gives  the  distribution  of  the  intensity  of 
the  field. 


PROBLEMS. 

1.  Find  the  potential  and  the  Held  intensity  due  to  a  hollow  sphere 
at  a  point  (I  i  outside,  (2)  within  the  hollow  part,  and  (3)  in  the  solid 
pari  of  the  Bphere. 

2.  Find  the  potential  and  the  field  intensity  due  to  a  circular  disk  of 
negligible  thickness  .-it  a  point  on  its  axis. 

3.  Find  the  potential  and  the  field  intensity  due  to  a  straight  wire  of 
length  /  and  mass  m  at  a  point  on  the  axis  of  the  wire.     The  cross-section 

negligible. 


FIELDS  OF  FORCE  AND  NEWTONIAN   POTENTIAL     217 

4.  Find  the  potential  and  the  field  intensity  due  to  a  straighl  circu- 
lar rod  at  a  point  on  its  axis. 

5.  Show  that  problems  2  and  3  arc  special  cases  of  problem  4. 

6.  Find  the  magnetic  potential  and  the  field  intensity  due  to  a,  cylin- 
drical magnet  at  a  point  on  its  axis;  suppose  the  magnetism  to  be  dis- 
tributed at  the  ends  only. 

7.  Find  the  potential  and  the  field  intensity  clue  to  two  spherical 
charges  at  a  point    equidistant  from  centers  of  the  two  charges. 

8.  Find  the  potential  and  the  field  intensity  due  to  a  right  cone  al  a 
point  on  its  axis. 

9.  A  uniform  solid  sphere  is  cut  in  two  by  a  diametral  plane.     Show 

3      tn 
that  the  gravitational  force  between  the  two  parts  will  be  —  y  — -,  where 

16     a2 

m  is  the  mass  of  the  sphere,  a  the  radius,  and  y  the  gravitational  constant. 

10.  Show  that  if  any  two  points  on  the  ■surface  of  the  earth  were  joined 
by  a  straight  and  smooth  tunnel  a  particle  would  traverse  it  in  about  42.5 
minutes. 

11.  Two  spheres  of  masses  m  and  m'  attract  each  other  with  a  force. 

F  =  y  — —  ,  where  y  is  a  constant  and  r  is  the  distance  between  the  centers. 

Taking  the  configuration  when  the  spheres  are  in  contact  to  be  that  of 
zero  potential  energy,  find  their  potential  energy  when  the  centers  are 
separated  by  a  distance  D.     The  radii  of  the  spheres  are  a  and  b. 

12.  In  the  preceding  problem  suppose  the  spheres  to  repel  each  other 
with  the  same  law  of  force  and  take  the  configuration  when  the  spheres 
are  separated  by  an  infinite  distance  to  be  that  of  zero  potential  energy. 

13.  Find  the  potential  due  to  a  small  magnet  at  a  point  whose  distance 
is  large  compared  with  the  length  of  the  magnet. 

14.  In  the  preceding  problem  find  the  components  of  the  intensity  of 
the  field  along  and  at  right  angles  to  the  line  joining  the  point  to  the 
magnet.     Also  find  the  total  intensity  and  its  direction. 


CHAPTER  XL 
UNIPLANAR   MOTION   OF   A   RIGID   BODY. 

181.  Angular  Kinetic  Reaction.  —  It  will  be  remembered 
thai  in  considering  the  equilibrium  of  rigid  bodies  the  Law 
of  Action  and  Reaction  was  divided  into  the  following  two 
sections: 

To  every   linear  action  there  is  an  equal   and   opposite 

linear    reaction,  or,  the   sum   of  all   the   linear  actions  to 

which  a  body  or  a  part  of  a  body  is  subject  at  any  instant 

vanishes.  .     . 

SA,  =  0.  (A,) 

To  every  angular  action  there  is  an  equal  and  opposite 
angular  reaction,  or,  the  sum  of  all  the  angular  actions  to 
which  a  body  or  a  part  of  a  body  is  subject  at  any  instant 

vanishes. 

^Au  =  0.  (AJ 

In  Chapter  VI  the  first  section  of  the  law  was  applied 
to  particles  in  motion;  but  in  order  to  do  this  the  meaning 
of  the  terms  "linear  action"  and  "linear  reaction"  was 
enlarged  bo  as  bo  include  linear  kinetic  reactions  as  well  as 
forces.  In  the  present  chapter  the  second  section  of  the 
law  will  be  applied  to  the  motion  of  rigid  bodies;  but  before 
doing  this  we  must  introduce  another  form  of  kinetic  re- 
action, which  we  will  call  angular  kinetic  reaction.  If  we 
replace  in  the  Becond  section  of  the  law  the  terms  "angu- 
lar  action"  and  "angular  reaction"  by  the  terms  "torque" 
and  "angular  kinetic  reaction,"  we  obtain  the  following 
form  which  is  directly  applicable  to  problems  of  rotation: 

The  sum  of  all  the  torques  acting  upon  a  rigid  body 
plus    the    angular    kinetic    reaction    equals    zero,    or   the 

218 


UXIPLAXAR  MOTION  OF  A  RIGID  BODY  219 

resultant  torque  equals  and  is  oppositely  directed  to  the 
angular  kinetic  reaction. 

Resultant  torque=  —(angular  kinetic  reaction).      (I) 

In  order  to  understand  the  nature  of  the  angular  kinetic 
reaction  consider  the  following  experiment:  If  we  try  to 
rotate  a  flywheel,  which  is  free  to  move  about  a  horizontal 
axis,  by  pulling  at  one  end  of 
a  string  which  is  wound  around 
the  axle,  Fig.  110,  we  find  that 
the  greater  the  angular  veloc-. 
ity  which  we  want  to  impart 
in  a  given  interval  of  time 
the  harder  we  must  pull  at 
the  string.     But  since  the  pull 

of  the  string  and  the  reaction  of  the  bearings  form  a  couple 
and  since  the  increase  in  the  angular  velocity  per  unit  time 
means  angular  acceleration,  we  conclude  that  a  torque  must 
be  applied  to  the  flywheel  in  order  to  impart  to  it  an  angular 
acceleration,  and  that  the  greater  the  acceleration  desired 
the  greater  must  the  torque  be.  Evidently  the  torque 
which  we  apply  to  the  flywheel  expends  itself  in  overcoming 
certain  reactions.  The  resisting  torque  due  to  the  friction 
between  the  axle  and  its  bearings  and  between  the  surface 
of  the  flywheel  and  the  surrounding  air  must  be  overcome. 
But  if  we  gradually  diminish  this  resisting  torque  by  reducing 
the  friction  we  observe  that  the  torque  which  must  be  applied, 
in  order  to  give  the  flywheel  a  certain  angular  acceleration, 
tends  towards  a  constant  value  different  from  zero.  In 
other  words  even  if  all  the  resisting  torques  due  to  friction 
were  eliminated  we  would  have  to  apply  a  torque  of  definite 
magnitude  in  order  to  give  the  flywheel  a  desired  angular 
acceleration;  that  is,  tin-  flywheel  resists  torques  which  im- 
part to  it  an  angular  acceleration.  This  resistance  to  angular 
acceleration  is  the  angular  kinetic  reaction. 


220  ANALYTICAL  MECHANICS 

182.  Experimental  Definition  of  Moment  of  Inertia.  —  If  in 
the  experiment  of  the  preceding  section  all  frictional  forces 
and  torques  are  eliminated  and  then  torques  of  different 
magnitude-  are  applied  to  the  flywheel,  it  will  be  found  that 
t  Ik-  t  <  >rques  are  proportional  to  the  angular  accelerations  pro- 
duced; that  is,  if  <;u  G2,  etc.,  denote  the  torques  obtained 
by  multiplying  the  pull  of  the  string  by  the  radius  of  the  axle 
and  yi,  yt}  etc.,  the  corresponding  angular  acceleration,  then 
we  shall  find  that  the  following  relations  hold: 

^  =  ^  =  ^=...=7,  (II) 

71        72        73 

where  /  is  a  constant  which  depends  only  upon  the  rotating 
Bystem.  In  fact,  as  will  be  shown  in  §  186,  it  is  nothing 
more  or  less  than  the  moment  of  inertia  of  the  rotating 
system.  We  have,  therefore,  the  following  definition  for 
the  moment  of  inertia  of  a  body,  in  addition  to  the  analytical 
definition  given  in  Chapter  VII: 

The  moment  of  inertia  of  a  body  about  a  given  axis  is  a  con- 
st <  ml  of  the  body,  relative  to  the  given  axis,  which  equals  the 
quotient  of  the  torque  applied  by  the  angular  acceleration  ob- 
tained; both  being  referred  to  the  given  axis* 

183.  Measure  of  Angular  Kinetic  Reaction.  —  It  is  evident 
from  equation  (II)  that  Gh  (?•>,  etc.,  which  measure  the  angu- 
lar kinetic  reactions  of  the  flywheel  for  the  accelerations  71, 
v..  He,  are  proportional  to  these  accelerations.  Therefore 
the  angular  kinetic  reaction  of  a  body  varies  directly  with 
the  angular  acceleration  imparted.  If,  on  the  other  hand,  a 
number  of  bodies  of  different  moments  of  inertia  are  given 
tin'  Bame  angular  acceleration,  it  is  found  that  the  kinetic  re- 
art  ion-  are  proportional  to  the  moments  of  inertia;  that  is, 

—  =  —  =•••=  7,  (HI) 

l\        12 

the  striking  similarity  between  this  definition  of  moment  of  inertia 
i'  definition  of  mass  given  in  §94. 


DNDPLANAB  MOTION  OF  A  RIGID  BODY  221 

where  y  is  the  common  angular  acceleration.  Therefore  the 
angular  kinetic  reaction  varies  directly  as  the  product  of  the 
moment  of  inertia  by  the  angular  acceleration, 

angular  kinetic  reaction  =  kly, 
where  k  is  the  constant  of  proportionality.     When  all  the 
magnitudes  involved  in  the  last  equation  are  measured  in 
the  same  system  of  units  k  becomes  unity. 

Introducing  this  simplification  in  the  last  equation  and 
putting  it  into  vector  notation  we  have 

angular  kinetic  reaction  =  — ZY.  (IV) 

The  negative  sign  indicates  the  fact  that  the  direction  of 
the  angular  kinetic  reaction  is  opposed  to  that  of  the  angular 
acceleration. 

184.  Torque  Equation.  —  Combining  equations  (I)  and  (IV) 
and  denoting  the  resultant  torque  by  G  we  obtain 

G  =  ?'}  (V) 

The  last  equation,  which  will  be  called  the  torque  equation, 
states  that  the  resultant  torque  about  any  axis  equals  the 
product  of  the  moment  of  inertia  by  the  angular  accelera- 
tion and  has  the  same  direction  as  the  angular  acceleration. 

185.  The  Two  Definitions  of  Moment  of  Inertia.  —  In  order 
to  show  that  the  constant,  I,  of  equation  (II)  and  the 
moment  of  inertia  defined  by  equation  (II)  of  page  152 
are  the  same  magnitude,  consider  the  motion  of  the  rigid 
body  A,  Fig.  Ill,  about  a  fixed  axis  through  the  point  0, 
perpendicular  to  the  plane  of  the  paper.  Let  dF  be  the 
resultant  force  acting  upon  an  element  of  mass  dm,  that 
is,  the  vector  sum  of  the  forces  due  to  externa]  fields  of 
force  and  the  forces  due  to  the  connection  of  dm  with  the 
rest  of  the  body.     Then 

at 
is  the  force  equation  for  the  element  of  mass. 


222 


ANALYTICAL  MECHANICS 


The  linear  acceleration  varies  from  point  to  point,  but 
the  angular  acceleration  is  the  same  for  all  the  elements. 
Therefore  the  discussion  of  the 
problem  becomes  simpler  if  we 
replace  the  linear  acceleration  by 
the  angular  acceleration.  This 
may  be  done  by  taking  the  mo- 
ments of  the  forces  about  the 
axis.  Since  dm  can  move  only 
in  a  direction  perpendicular  to 
the  line  r,  the  resultant  force  d? 

must  be  perpendicular  to  r.     Therefore  the  magnitude  of 
the  moment  dG,  due  to  dF,  is 
dG=rdF 

dv 


Fig.  111. 


=  rdm 


dt 


=  r2  dm 


dt 


(v  =  rco). 


Therefore  the  resultant  torque  acting  upon  the  body,  or  the 
Bum  of  the  moments  due  to  the  forces  acting  upon  all  the 
particles  of  the  body,  is 

=  co  I    r2  dm. 

Bui  by  equation  (V)  G  =  Iw.    Therefore 
/=  /    r2dm,    ' 

which  i-  the  definition  of  the  moment  of  inertia  given  in 
Chapter  VII. 

186.    Comparison. — There  is  a   perfect  analogy  between 
D    of  pure  translation  and  motion  of  pure  rotation. 
This  is  clearly  brought    out   in  the  following  lists  of  the 
magnitudes  involved  in  the  two  types  of  motion: 


UXIPLAXAR  MOTION  OF  A  RIGID  BODY 


223 


Magnitudes  involved  in  motion  of  translation. 

Their  analogues  in  motion  of  rotation. 

s,  linear  displacement. 

8,  angular  displacement. 

v,  lincai-  velocity. 

«),  angular  velocity. 

v,  linear  acceleration. 

(I),  angular  acceleration. 

///,  linear  inertia  or  mass. 

/,  angular  inertia  or  moment  of 

inertia. 

—  mv,  linear  kinetic  reaction. 

—  7o>,  angular  kinetic  reaction. 

///V,  linear  momentum.* 

Ica,  angular  momentum,  f 

.1  mv2,  kinetic  energy  of  translation. 

2-  loo2,  kinetic  energy  of  rotation. 

F  =  mv,  force  equation. 

G  =  /<5),  torque  equation. 

1T=  C  F  ds,  work  done  bv  a  force. 

Jo 

W=  (   G  do,  work  done  by  a  torque. 

Jo 

L  =  (   F  dt,  linear  impulse.* 

Jo 

H  =  C  Gdt,  angular  impulse,  t 

Jo 

*  Discussed  in  Chapter  XII. 


t  Discussed  in  Chapter  XIII. 


187.  Torque  and  Energy  Methods.  —  The  equations  of 
motion  of  a  rigid  body  may  be  obtained  in  two  ways,  one 
of  which  will  be  called  the  torque  'method  and  the  other  the 
energy  method. 

Torque  method:  First,  find  the  resultant  torque  and  sub- 
stitute it  in  the  torque  equation. 

Second,  integrate  the  torque  equation  in  order  to  find 
the  integral  equations  of  the  motion. 

Energy  method:  First,  equate  the  change  in  the  potential 
energy  to  the  change  in  the  kinetic  energy. 

Second,  differentiate  the  energy  equation,  thus  obtained, 
with  respect  to  the  time.     This  gives  the  torque  equation. 

Third,  proceed  as  in  the  torque  method. 

The  energy  method  is  advantageous  in  complicated  prob- 
lems, but  not  in  simple  ones. 


224  ANALYTICAL  MECHANICS 

ILLUSTRATIVE   EXAMPLES 
ON    THE  MOTION   OF  A   RIGID  BODY  ABOUT    A  FIXED  AXIS. 

Discuss  the  motion  of  a  rigid  body,  which  is  free  to  rotate  about  a  fixed 
axis,  under  the  action  of  a  constant  torque. 

Suppose  the  body  to  be  the  flywheel  of  Fig.  110.  Let  the  constant 
torque  be  supplied  by  a  constant  force  F  applied  at  the  free  end  of  the 
string  which  is  wound  around  the  axle.  The  tensile  force  of  the  string 
and  the  reaction  of  the  bearings  form  a  couple,  the  torque  of  which  equals 
the  moment  of  the  force  F  about  the  axis  of  rotation.  Therefore 
G  =  Fa, 

where  a  is  the  radius  of  the  axle.     Substituting  this  value  of  G  in  the  torque 
equation  we  obtain 

7  "77    =  Fa> 
(It 

r/co       Fa 
or  —  =  —  =  7  =  const. 

[ntegrating  the  last  equation  we  get 

co  =  yt  +  c. 
Let  co  =  co0  when  t  =  0,  then  c  =  co0.     Therefore 

CO   =  C00  +  (lit,  (1) 

(Id 

or  —   =  co0  +  oot. 

at 

[ntegrating  again 

8  =  oo0t  +  J  cot2  +  c'. 

Let  6  =  0  when  t  =  0,  then  c'  =  0.     Therefore 

e  =  u0t  +  i  co/-.  (2) 

Eliminating  /  between  equations  (1)  and  (2) 

co';  =  coo2  +  270.*  (3) 

I  \i  i;..y   Method. — The  increase  in  the  kinetic  energy  due  to  the 
action  of  F  is 

T  -T0  =  ]  /or  -  J  /coo2. 

The  diminution  in  the  potential  energy  of  the  system  which  supplies  the 
F  equals  the  work  done  by  F.    Therefore 

-(U-  U0)  =  Fs, 

Compare  equations  (1),  (2),  and  (3)  with  the  corresponding  equations 
of  p.  113. 


UNIPLANAR  MOTION  OF  A  RIGID  BODY 


225 


where  s  is  the  length  of  the  string  which  is  unwound.    Substituting  these 
in  the  energy  equation  we  obtain 

$  Zco2  —  5  7co02  =  Fs. 

op 

.*.      or  =  co02  +    ^y-  s 

=  w02 +  2  70,     (s  =  ad) 

which  is  the  equation  (3)  obtained  by  the  torque  method.     Differentiating 
the  last  equation  we  have 

2co^  =  27co 
at 

dco 

dt 


and 


7, 


which  is  the  equation  obtained  by  the  torque  method;  therefore  the  rest 
of  the  problem  is  identical  with  that  given  by  the  torque  method. 

2.  A  flywheel  rotates  about  a  horizon- 
tal axis  under  the  action  of  a  falling  body, 
which  is  suspended  by  means  of  a  string 
wound  around  the  axle  of  the  flywheel. 
Discuss  the  motion,  neglecting  the  mass  of 
the  string. 

Let      /  =  the  moment  of  inertia  of  the 
rotating  system. 
m  =  the  mass  of  the  falling  body. 
a  =  the  radius  of  the  axle. 
T  =  the  tensile  force  of  the  string. 


Torque  Method.  —  Taking  the 
moments  about  the  axis  of  rotation 
we  have  q  _  m_ 

for  the  resultant  torque.  There- 
fore  la  =  Ta 

is  the  torque  equation.  But  con- 
sidering the  forces  acting  upon  the 
falling  body  we  gel 

mi'  —  mg  —  T. 
Hence 

16)  =  Ta 

=  (mg  —  mv)  a 
=  m  (g  —  acb)  a. 


Energy  Method. — Suppose  the 
flywheel  to  start  from  rest,  and  let  h 
denote  the  distance  covered  by  the 
body  during  its  fall.  Then  the 
energy  equation  gives 

J  /or  +  \  mv'1  =  mgli. 

Differentiating  with  respect  to  t, 

Juiio  -\-  mm  =  mgh. 
Hut  V  =  au  and  li  =  i\  therefore 

/cow  -+-  /wi'-coco  =  mgdST, 
or  I co  =  m  ((/  —  au 


226 


ANALYTICAL  MECHANICS 


Thus  we  have  from  either  method 


/  +  ma* 


I  +  ma2"' 
It  is  evident  from  the  last  two  equations  that  both  the  linear  and  the 
angular  accelerations  are  constant.     Therefore  the  equations  of  motion 
are 


/  +  ma 

1  ma2 

2  1  +  ma 


gt, 


,gt2, 


i'2  =  2 


/  +  ma2 
1        ma 


I  +  ma1     ' 

Discussion.  —  When  /  <C  ma,  then 
v  =  g,  and  the  motion  of  the  sus- 
pended body  is  about  the  same  as 
thai  of  a  freely  falling  body.  When 
I  ^>  ma  then  v  =  0.  Therefore  the 
velocity  of  the  falling  body  changes 
very  Blowly. 

3.   A  uniform  rectangular  trapdoor, 

which  is  held  in  a  vertical  position,  is 
allowed  to  fall.  Supposing  the  hinges 
In  be  smooth  and  horizontal,  find  the 

expre—iun  for  the  angular  velocity  a1 

any  instant  of  the  motion. 

Toeqi  i.  Method. —  The  torque 
on  the  door  is  due  to  the  action  of 
it-  weight  and  the  reaction  of  the 

Therefore 


2/  +  mw 

2       ma     , 
I  +  ma2' 


gt\ 


Fig.  113. 


' 


sin 


Putting   this    value   of   G    in    the 

torque  equation  we  gel 

Zw.^sin*. 

- 


Energy  Method.  —  In  turning 
through  an  angle  6  the  door  acquires 
a  kinetic  energy  of  '.  Ztts  and  loses 
from  its  potential  energy  an  amount 
equal  tomgh.  Therefore  the  energy 
(•(liiation  gives 


Iuy  =  mgh 


>»g  o  (1 


cos  6). 


I  differentiating 
the  time 


lit 


with  respect  to 


UXIPLAXAR  MOTION  OF  A  RIGID  BODY  227 


•'•    w  =  iff sin  R, 

dft 

Multiplying  both  sides  of  the  last  equation  by  2  —  dt,  and  integrating  we 

obtain 

3.9 


(f)" 


cos  0  +  c. 


But  ^  =  0  when  0  =  0,  therefore  c  =  ^.     Hence 
dt  a 

W2  =  3-2(1  -cos0). 
a 

Discussion.  —  It  will  be  observed  that  this  result  is  already  given  by 
the  energy  equation. 

When  6=  -,  co2  =  — ,  therefore  the  door  strikes  the  floor  with  an 
2  a 

angular  velocity  of  i  /zJL .     Thus  the  greater  a  the  less  the  angular  velocity 

V    a 
with  which  the  door  strikes  the  floor.     On  the  other  hand  the  linear 
velocity  with  which  the  end  of  the  door  strikes  the  floor  increases  with 
a,  since 

v2  =  Sag  (1  —  cos0) 

=  3  ag,  when  6  =  -  • 

PROBLEMS. 

1.  Discuss  the  motion  of  the  falling  bodies  in  Atwood's  machine, 
supposing  the  pulley  to  rotate  without  slipping. 

2.  In  the  problem  discussed  in  the  first  illustrated  example  take  into 
account  the  resistance  of  the  air,  supposing  the  resistance  to  be  propor- 
tional to  the  angular  velocity  of  the  wheel. 

3.  A  flywheel  which  is  making  loo  revolutions  per  minute  and  which  is 
subject  to  a  constant  torque  of  50  pounds-fool  comes  to  resl  after  making 
1500  revolutions.  Find  the  moment  of  inertia  of  the  wheel,  the  angular 
acceleration  and  the  time  taken  incoming  to  rest.  The  angular  accelera- 
tion is  supposed  to  be  constant. 

4.  A  flywheel  which  is  sub j eel  to  a  constant  torque  of  5000  dynes- 
centimeter  stai-ts  from  resl  and  makes  2000  revolutions  in  1  minutes. 
Find  the  angular  acceleration  and  the  moment  of  inertia. 


228  ANALYTICAL  MECHANICS 

5.  In  the  Becond  illustrative  problem  suppose  there  is  a  resistance  to 
the  motion  of  the  falling  body  proportional  to  its  velocity. 

6.  A  flywheel  making  400  revolutions  per  minute  is  brought  to  rest  in 
3  minutes  by  means  of  friction  brakes  applied  to  it.  Find  the  angular 
inertia  of  the  wheel  and  axle  if  the  total  brake-shoe  force  applied  is  500 
pounds  and  the  diameter  of  the  flywheel  is  10  feet. 

7.  Int  he  preceding  problem  find  the  total  number  of  revolutions  made 
after  the  brakes  were  applied. 

8.  A  flywheel  is  brought  to  rest  by  means  of  brakes  applied  at  the 
axle.  If  the  combined  angular  inertia  of  the  flywheel  and  the  axle  is 
50,000  gm.  cm.2  and  the  diameter  of  the  axle  20  cm.,  find  the  force  which 
must  bo  applied  on  the  brakes  in  order  to  bring  the  flywheel  to  rest  within 
5  minutes,  the  initial  angular  velocity  being  30  radians  per  sec. 

9.  A  flywheel  is  stopped  by  fluid  friction.  The  resisting  torque  due  to 
the  friction  is  proportional  to  the  angular  velocity.     Discuss  the  motion. 

10.  The  flywheel  of  a  gyroscope  is  rotated  by  applying  a  force  to  a 
string  wound  around  the  axle.  Discuss  the  motion,  supposing  the  tension 
of  the  string  to  be  proportional  to  the  length  of  the  string  unwound. 

MOTION  OF  A  RIGID  BODY  ABOUT  INSTANTANEOUS  AXES. 

188.  Uniplanar  Motion.  —  It  was  shown  on  p.  31  that  uni- 
planar  motion  may  be  considered  as  a  motion  of  pure  rotation 
at  each  instant  of  the  motion.  Since  the  torque  and  energy 
equations  hold  good  at  each  instant  of  the  motion  they  can 
be  applied  to  a  rigid    body  in 

uniplanar  motion  as  if  the  in-  0*s 
Mantaneous  axis  were  fixed  at 
the  instant  considered.  There- 
fore uniplanar  motion  may  be 
discussed  in  the  same  way  as 
iimt  ion  about  a  fixed  axis. 

189.  Instantaneous  Axis.  —  If 
at  any  instant  tbe  velocities  of 
two  points  of  a  rigid  body  are 
known   the   position  of   the  in- 

Btantaneous  axis  may  be  found  in  the  following  manner: 
Lei  /'  and  Q,  Fig.  Ill,  be  two  points  which  lie  in  a  plane 
parallel  to  the  guide  plane,  and  the  velocities  of  which  are 


UNIPLAXAR  MOTION'  OF  A  RIGID  BODY 


229 


not  parallel;  further  let  vP  and  wQ  be  the  velocities.  Draw 
the  line  PO  perpendicular  to  vP  in  a  plane  parallel  to  the 
guide  plane;  also  draw  QO  perpendicular  to  vQ  in  the  same 
plane.  Then  the  instantaneous  axis  passes  through  0,  the 
point  of  intersection,  and  is  perpendicular  to  the  plane  POQ. 

ILLUSTRATIVE  EXAMPLES. 
1.    Discuss  the  motion  of  a  uniform  circular  cylinder  which  rolls  down 
a  rough  inclined  plane  without  slipping. 


Fig.  115. 

Let  m  =  the  mass  of  the  cylinder. 

I  =  the  moment  of  inertia  of  the  cylinder  about  the  element  of 

contact. 
a  =  the  radius  of  the  cylinder. 
v  =  the  velocity  of  the  axis  of  the  cylinder, 
w  =  the  angular  velocity  of  the  cylinder. 


Torque  Method.  — The  torque 
is  due  to  the  weight  of  the  cylinder 
and  the  reaction  of  the  plane.  It 
equals  the  moment  of  the  weight 
about  the  element  of  contact. 
Therefore 

G  =  mga  sin  a. 
Substituting  this  value  of  G  in  the 
torque  equation  we  have 
I  Co  =  mga  sin  a. 


Energy  Method.  —  In  moving 
through  a  v  distance  s  aloii.ti  the 
plane  the  potential  energy  of  the 
cylinder  is  diminished  by  an  amount 
equal  to 

mgh  =  mgs  sin  <>. 
Therefore  the  energy  equation  gives 

\Io)2  -  \  Iu)02  =  »i0ssin^ 
Differentiating  with  respect  to  the 
time  ^  |    :', 

Iuioj  =  mgs  sin  a. 
:.     J(jj  =  mga  sin  a. 


230 


ANALYTICAL  MECHANICS 


Therefore 
and 


w  =  — -  g  sin  a, 
6a 


Thus  both  the  linear  and  the  angular  accelerations  of  the  cylinder  are 
constant.    Therefore  the  equations  of  the  motion  are  the  following: 


v  =  Vo  +  ^gtana, 

s  =  v0t  +  -  gt2  sin  a, 

o 

4 

''2  =  <'<r  +  ^  gs  sin  a. 
o 


co  =  co0  +  tt~  g  sin  a, 
Sa 

6  =  co0t  +  —  gt2  sin  a, 

4 
co2  =  oj02  +  —  gQ  sin  a. 
3a 


2.    A  wheel  moves  down  an  inclined  groove  with  its  axle  rolling  along 
the  groove  without  slipping.     Discuss  the  motion. 
Let  a  =  the  radius  of  the  axle. 
b  =  the  radius  of  the  wheel. 
m'—  the  mass  of  that  part  of  the 
axle  which  projects  out  from 
the  wheel. 
m  =  the  mass  of  the  rest  of  the 

moving  system. 
.1/  =  m  +  ///'. 


Fig.  116. 


Suppose  the  wheel  to  be  a  solid  disk 
with  a  thickness  equal  to  half  the  total 
length  of  the  axle.    Then  if  both  the  wheel  and  the  axle  are  of  the  same 


material  the  relation 


lr 


holds.     Therefore 


M    and 


a2 +  6 

Tobqub  Method. — Considering 

the  momenta  about  the  element  of 
contacl  we  obtain  the  following  for 
the  torque  equation: 

Iu>  =  Mga  sin  a, 
(Ic+Mn '■'>  u      Mga  sin  a, 
(Ie  +  Ma7)  v  =  Mga*  sin  a, 
'     ':■  QOtl  3  thfi  moment  of  in- 
"f  the  moving  system  about 


M. 


a2  +  6" 

Energy  Method.  —  Supposing 
the  wheel  to  start  from  rest  we  ob- 
tain 

'.  Mv*  -f  I  7coj2  =  Mgs  sin  a, 

i  I  Mr-  +  Ic  —  I  =  Mgs  sin  a, 

\  (Ic  +  Ma2)  v2  =  Mga2s  sin  a, 
(Ic  +  Ma2)  w  =  Mga2s  sin  a, 
(Ie  +  Ma2)  i)   =  Mga2  sin  a. 


UNIPLANAR  MOTION  OF  A  RIGID  BODY 


231 


Ma* 


Ma  . 

•'•     (a  =  l^Ma->gSma- 

Thus  both  the  linear  acceleration  and  the  angular  acceleration  are 
constant.  Therefore  the  equations  of  motion  may  be  obtained  as  in  the 
preceding  problem. 


Discussion. 


/«  = 


!  .   mb2 
'"•"    2 
a*  +  b* 


2  (a2  +  b2) 
Substituting  this  value  of  Ic  in  the  expression  for  v  we  get 


.  2  a2  (a2  +  b2) 


,  g  sin  a. 


2a4  +  (a2  +  62)s 

Case  I.  —  Let  b  =  a,  then  v  =  \  g  sin  a,  which  is  the  acceleration  of 
a  cylinder  rolling  clown  an  inclined  plane. 
2 


Case  II. — Let6 


then 


Caselll.  —  Letfr^>«,  then 


g  sin  a,  as  in  case  I. 

'  sin  a.     Thus  by  reducing  the 

radius  of  the  axle  we  can  reduce  the  acceleration,  theoretically  at  least,  as 
much  as  we  please.  The  reason  for  this  fact  becomes  clear  when  we  con- 
sider the  relative  proportions  of 
the  potential  energy  which  are 
transformed  into  kinetic  energy 
of  translation  and  kinetic  energy 
of  rotation. 

3.  In  Fig.  117  the  larger  cir- 
cle represents  a  cylinder  of  mass 
M  which  rolls  along  a  rough  hori- 
zontal table,  under  the  action 
of  a  falling  body  of  mass  m.  The 
right-hand  end  of  the  ribbon, 
which  connects  the  falling  body 
with  the  cylinder,  is  wound 
around  the  latter  so  that  it  is 
unwound  as  the  motion  goes 
on.  The  pulley  over  which  the 
ribbon  slides  is  smooth.  Discuss  the  motion,  supposing  the  mass  and 
the  thickness  of  the  ribbon  to  be  negligible. 


232  ANALYTICAL   MECHANICS 

Torqik  METHOD.  —The  cylinder  is  acted  upon  by  four  forces — its 

weight  .Ug,  the  aorrnal  reaction  N,  the  frictional  reaction  F,  and  the 

tensile  force  of  the  string  T.    Taking  the  moments  about  the  element  of 

contact  we  obtain 

G  =T.2a, 

where  n  is  the  radius  of  the  cylinder.  The  other  forces  do  not  have 
mom<  nts  about  the  element  of  contact.  But  considering  the  motion  of 
the  falling  body  we  find  that 

mv  =  mg  —  T, 

where  b  is  the  acceleration  of  the  falling  body.     Therefore 

G  =  m  (g  —  b)  •  2  o. 
Substituting  this  value  of  G  in  the  torque  equation, 

7o>  =  2  aw  (a  —  b), 
where  I  is  the  moment  of  inertia  of  the  cylinder  about  the  element  of  con- 
tact,  and  u  the  angular  acceleration.  But  since  the  highest  element  of 
the  cylinder  has  the  same  linear  velocity  as  the  ribbon  and  the  falling  body, 
we  have  _»  ou  =  v,  and  consequently  cb  =  ~  .  Making  this  substitution 
in  the  torque  equation 

^a  =  2am(g-v), 
4  a^rn 


7  +  4  a*m 
m 


(I  =  I  Ma-.) 


m+lM  " 

].m.i;<;y  Method.  —Supposing  the  initial  velocities  to  be  zero  and 
equating  the  gain  in  the  kinetic  energy  of  the  system  to  the  loss  in 
potential  energy  we  have 

J  r«2  +  irrw!  =  mgh, 

where  h  i<  the  distance  Mien  through  1 1 y  the  body.     Differentiating  the 
lasl  equation  with  respect  to  the  time, 

/cow  +  mob  =  mgh. 

But  A  —  V.  w  «=  --—  ,  andeb  =  ^—.  Making  these  changes  and  solving  for 

■l  a  2  a 

b  we  obtain 

•  •_     _1^L_  •  _      2fl»t 

"i  +  iahn9  "      I  +  la-m0, 

m  m 


m+W  'la\m+\M) 

which  an-  the  expressions  obtained  by  the  torque  method. 


UNIPLANAR  MOTION  OF  A  RIGID  BODY  233 

Discussion.  —  It  is  evident  that  both  b  and  u  arc  constant.  There- 
fore the  motions  of  both  the  cylinder  and  the  falling  body  are  uniformly 
accelerated,  the  one  in  rotation,  the  other  in  translation. 

When  m  is  negligible  compared  with  M,  v  is  very  small  and  consequently 
the  motion  very  slow.  When  m  is  very  large  compared  with  M,  v  is  prac- 
tical^ equal  to  g,  hence  the  body  falls  almost  freely. 

The  linear  acceleration  of  the  axis  of  the  cylinder  equals  one-half  that 
of  the  falling  body.  The  linear  accelerations  of  the  cylinder  and  of  the 
falling  body  depend  upon  the  radius  of  the  cylinder  only  indirectly,  i.e., 
through  the  mass  of  the  cylinder. 

4.  A  circular  hoop  is  projected  along  a  rough  horizontal  plane  with  a 
linear  velocity  v0  and  an  angular  velocity  co0.     Discuss  the  motion. 

The  hoop  is  acted  upon  by  two  forces,  namely,  its  weight  and  the  re- 
action of  the  plane.  The  latter  may  be  resolved,  as  usual,  into  its  normal 
component  N  and  its  frictional  component  F.  Then  the  force  equation 
gives 

.*-*!■  (!) 

for  the  horizontal  direction  and 

0  =  N  -  mg  (2) 

for  the  vertical  direction.     On  the  other  hand  the  torque  equation  gives 

Mh**  (3) 

where  a  is  the  radius  of  the  hoop  and  Ic  its  moment  of  inertia  about  its 
own  axis.  The  double  sign  indicates  the  fact  that  the  direction  of  F 
changes  with  the  direction  in  which  slipping  takes  place  at  the  poinl  of 
contact.  Denoting  the  coefficient  of  friction  at  the  point  of  contact  by  fx 
we  have 

F  =  imN, 
=  fxmg  [by  equation  (2)]. 

Making  this  substitution  in  equations  (1)  and  (3)  and  replacing  Ie  in 
equation  (3)  by  its  value  we  obtain 

!=*«  <4> 

*-*«,  (5, 

Case  I.  —  Suppose  the  initial  angular  velocity  to  be  clockwise  and 


234  ANALYTICAL  MECHANICS 

v0  <  acfe  Then  the  sliding  at  the  point  of  contact  is  toward  the  left; 
therefore  F  is  directed  to  the  right  and  consequently  positive.    Thus 

dt  =  txg>  (4) 

%         &L.  (5') 

at  a 

Integrating  the  last  two  equations  we  have 

v  =  v0  +  ngt.  (6) 

u>  =  u>o-^t.  (7) 

a 

These  equations  hold  until  sliding  stops,  after  which  the  hoop  rolls  with 
constant  angular  and  linear  velocities.  Let  h  denote  the  time  when 
sliding  -tops,  that  is,  when  v  =  aoo.    Then 


v0  +  ugh 


'(•*-£> 


au0  —  vp  /« 

or  ti  =  — - — —  (o) 

2  W 

Substituting  this  value  of  t  in  equations  (6)  and  (7)  we  get 

*  -  *±»,  (9) 

and  o>,  =  *|^4,  (10) 

for  the  linrai  and  the  angular  velocities  of  the  hoop  after  the  instant 
when  the  sliding  ceases.  The  subsequent  motion  is  one  of  pure  rolling 
with  a  linear  velocity  >i,  greater  than  v0,  and  angular  velocity  «i,  less 
than  co0. 

Case  II.  —  Initial  rotation  clockwise  and  Vo  >  aa)0-     In  this  case  slid- 
ing  i-  toward  the  right,  consequently  F  is  negative  and  therefore 

Tr- m,  (4) 

at        a 

If  I-  denotes  the  time  when  sliding  stops,  in  this  case,  a  reasoning  similar 
to  i he  foregoing  Lri\es 

U  =  ?-^F^>  (8') 

v,  =  ?^p,  (9') 

a,,  =  Sdt5».  (10') 


UNIPLAXAR  MOTION  OF  A  RIGID  BODY  235 

Therefore  after  the  time  t»  the  hoop  will  roll  along  towards  the  right  with 
a  linear  velocity  v2  less  than  rn,  and  with  an  angular  velocity  u  greater 
than  Wo. 

Case  III.  —  Suppose  the  initial  rotation  to  be  counter-clockwi.se.     In 

this  case  we  obtain 

5— #»  (i  ) 


dt~~  a 


(5'") 


fc=3L±«*  (n 


2ng 

v0  —  aoio 
2       ' 


(9") 


Vo  —  QUO  nr,l\ 

"3=^a-'  (1G) 

where  t3  is  the  time  when  sliding  ceases. 

There  are  three  special  cases  to  be  considered  : 

(a)  When  v0  >  aa>0,  v3  is  positive,  and  consequently  the  hoop  goes  on 
rolling  towards  the  right. 

(b)  When  v0  =  aa)0,  v3  =  0,  and'  consequently  at  t  =  t3  the  hoop  comes 
to  rest. 

(c)  When  v0  <  aai0,  v3  is   negative.     Therefore   at  the  instant  t  =  t3 
the  hoop  begins  to  roll  backwards. 

PROBLEMS. 

1.  Discuss  the  motion  of  the  following  bodies  rolling  down  an  inclined 
plane  without  slipping: 

(a)  A  hollow  cylinder  of  mass  m  and  inner  and  outer  radii  r{  and  r») 
respectively. 

(b)  A  hoop  of  mass  ///  and  radius  r. 

(c)  A  sphere  of  mass  m  and  radius  r. 

(d)  A  hollow   sphere  of  mass  m   and  inner  and  outer  radii  r,  and  r«, 

respectively. 

(e)  A  spherical  shell  of  negligible  thickness  of  mass  m  and  of  radius  r. 

(f)  Compare  the  time-;  of  descent   in  (c)  and 

2.  A  sphere  is  projected,  wiihout  initial  rotation,  up  a  perfectly  rough 
inclined  plane.      Discuss  the  motion. 

3.  A  wheel  which  is  rotating  aboul  its  own  axis  is  placed  on  a  per- 
fectly rough  inclined  plane.       DisCUSS  the  motion  up  the  plain-. 


236 


ANALYTICAL  MECHANICS 


4.  The  return  trough  of  a  howling  alley  is  50  feet  long  and  has  a  slope 
of  1  foot  in  20  feet.  Supposing  the  contact  to  be  perfectly  rough  find  the 
time  a  ball  will  take  to  return.     The  sides 

of  the  trough  are  perpendicular  to  each 
other. 

5.  In  the  adjoining  figure  the  largest  cir- 
cle represents  a  solid  disk  wheel,  which  rolls 
along  a  rough  horizontal  table  under  the 
action  of  a  falling  body.  The  left-hand  end 
of  the  string  is  spliced  and  connected  to  two 
smooth   rings  on   the   axle   of   the   wheel. 

The  pulley  over  which  the  string  passes  is  smooth.     Discuss  the  motion. 

6.  In  the  preceding  problem  suppose  the  pulley  to  be  rough  and  to 
rotate  about  its  axis. 

7.  Same  as  problem  5  except  that  the  wheel  rolls  up  an  inclined 
plane. 

8.  In  the  preceding  problem  suppose  the  pulley  to  rotate. 

9.  Same  problem  as  the  third  illustrative  example,  p.  231,  except  that 
the  cylinder  is  hollow  and  has  a  negligible  thickness. 

10.  Same  as  the  preceding  problem,  but  the  cylinder  rolls  up  an 
inclined  plane. 

11.  I  low  can  you  tell  a  solid  sphere  from  a  hollow  one  which  has  exactly 
the  same  diameter  and  mass? 

12.  Two  men  of  different  weights  coast  down  a  hill  on  exactly  similar 
bicycles.  Which  will  reach  the  bottom  of  the  hill  first,  the  lighter  or 
the  heavier  man? 

13.  A  thin  spherical  shell  of  perfectly  smooth  inner  surface  is  filled 
with  water  and  allowed  to  roll  down  an  inclined  plane.  Discuss  the 
motion. 

14.  A  hollow  cylinder  of  negligible  thickness  and  perfectly  smooth 
inner  surface  is  filled  with  water  and  allowed  to  roll  down  an  inclined 
plane.     Discuss  the  motion. 


GENERAL   PROBLEMS. 

1.  A  sphere  of  radius  a  starts  from  the  top  of  a  fixed  sphere  of  radius  b 
and  rolls  down.  If  there  is  no  sliding  find  the  position  at  which  they  will 
separate. 

2.  Two  masses  ///,  and  m%  are  suspended  by  means  of  strings  which  are 
wound  around  a  wheel  and  its  :i\le,  respectively.  The  wheel  and  axle 
are  rigidly  connected  and  are  free  to  rotate  about  a  horizontal  axis.    Dis- 

B  motion, 


UNIPLANAB  MOTION  OF  A  RIGID  BODY  :'}7 

(a)  When  .l/i  and  M:,  the  masses  of  the  wheel  and  axle,  are  negligible; 

(b)  When  they  are  not  negligible. 

3.  In  the  At  wood  machine  problem  show  that  if  the  pulley  is  not 

rough  enough  the  acceleration  of  the  two  moving  masses  is  -  —  a 

M  +  wie**  ' 
where  ju  is  the  coefficient  of  friction. 

Hint.  —  If  T  and  T'  are  the  tensile  forces  in  the  string  on  the  two  Bides, 
T=  Tel". 

4.  Same  as  the  third  illustrative  problem,  but  the  pulley  P  is  supposed 
to  rotate. 

6.  In  the  preceding  problem  suppose  the  cylinder  to  roll  up  an  inclined 
plane. 

6.  A  tape  of  negligible  mass  and  thickness  is  wound  around  the  middle 
of  a  cylinder.  The  free  end  of  the  tape  is  attached  to  a  fixed  point  and 
then  the  cylinder  is  allowed  to  fall.  Show  that  the  cylinder  falls  with  an 
acceleration  of  §  g  and  the  tensile  force  of  the  tape  is  1  W,  where  W  is  the 
weight  of  the  cylinder. 

7.  In  the  preceding  problem  the  fixed  point  is  on  an  inclined  plane 
and  the  cylinder  rolls  down  the  plane. 

8.  Discuss  the  motion  of  a  log  which  moves  along  its  length  down  an 
inclined  plane,  upon  two  rollers,  which  stay  horizontal. 

9.  A  uniform  rod  is  allowed  to  fall  from  a  position  where  its  lower  end 
is  in  contact  with  a  rough  plane  and  it  makes  an  angle  a  with  the  horizon. 

Show  that  when  it  becomes  horizontal  its  angular  velocity  is  y-^sin  a, 

where  I  is  the  length  of  the  rod. 

10.  Discuss  the  motion  of  a  cylinder  down  an  inclined  plane,  supposing 
the  contact  to  be  imperfectly  rough,  so  that  the  cylinder  both  slides  and 
rolls. 

11.  In  the  preceding  problem  suppose  the  cylinder  to  be  hollow. 


CHAPTER   XII. 
IMPULSE    AND    MOMENTUM. 

190.  Impulse. — It  was  stated  at  the  beginning  of  Chapter 
VIII  that  when  a  force  acts  upon  a  body  two  entirely  dif- 
ferent mechanical  results  are  produced  which  are  called  work 
and  impulse.  The  former  is  the  result  of  the  action  of  force 
in  space.  The  latter  is  the  result  of  the  action  of  force  in 
time.  We  have  already  discussed  work.  Impulse  is  the 
subject  of  the  present  chapter. 

191.  Measure  of  Impulse.  —  If  a  force  which  is  constant 
both  in  direction  and  magnitude  acts  upon  a  particle  the 
impulse  which  it  imparts  to  the  particle  equals  the  product 
of  the  force  by  the  time  during  which  it  acts.  Since  time  is 
a  scalar  while  force  is  a  vector,  impulse  is  a  vector  which 
has  the  same  direction  as  the  force.  If  L  denotes  the  im- 
pulse  which  a  constant  force  F  imparts  in  the  interval  of 
time  /,  we  can  write 

L  =  F  .  I  (I') 

When  the  force  is  variable  in  magnitude  or  in  direction,  or 
in  both,  we  must  consider  the  impulses  imparted  in  infini- 
tesimal intervals  of  time  and  add  them  up.     Thus 
dl_  =  Fdt 

and  '  L=  ffdt.  (I) 

Substituting  in  the  last  equation  mv  for  F  we  have 

L  =  /  mvdt 

Jo 


i  I  dv, 


/// 
=  mv  —  mvo,  (II) 

238 


IMPULSi:  AM)  MOMENTUM  239 

where  v0  and  v  are  the  velocities  at  the  instants  t  =  0  and 
t  =  t,  respectively.  If  v0  and  v  are  parallel,  equation  (II )  may 
be  written*  in  the  form 

L=  mv  —  mv0.  (II') 

192.  Momentum. — The  vector  magnitude  mv  is  called  mo- 
mentum. Therefore  the  momentum  of  a  particle  equals  the 
product  of  the  mass  by  the  velocity  and  has  the  same  direc- 
tion as  the  latter.  Equation  (II)  states,  therefore,  that  im- 
pulse equals  the  vector  change  in  momentum. 

PROBLEM. 

Show  that  the  component  of  the  impulse  along  any  direction  equals 
the  change  in  the  component  of  the  momentum  along  the  same  direction, 
that  is, 


S. 


X  dt  =  mx  —  mxn,  etc. 


193.    Dimensions  and  Units. — Substituting  the  dimensions 

of  force  and  time  in  the  definition  of  impulse  and  those  of 

mass  and  velocity  in  the  definition  of  momentum,  we  obtain 

[MLT-1]  for  the  dimensions  of  both.     The  C.G.S.  unit  of 

£m.  cm. 
both  impulse  and  momentum  is  the  —        — .     The  British 
1  sec. 

unit  is  the  pound-second. 

Force  and  Momentum.  —  Let  F  denote  the  resultant  of  all 
the  forces  acting  upon  a  particle  of  mass  m.     Then  we  have 
F  =  mv 

=  |(mv),  (III) 

which  states  that  the  resultant  force  experienced  by  a  particle 
equals  the  time  rati'  of  change  of  the  momentum  of  the  particle. 
In  order  to  extend  this  result  to  a  system  of  particles  lei 
F  denote  the  resultanl  of  all  the  external  forces  acting  upon 
the  system.  Further  let  F,  be  the  resultanl  of  all  the  forces 
acting  upon  any  one  of  the  particles.  Evidently  F  is  the 
resultant  of  two  sets  of  forces,   namely,   those  which  are 


240  ANALYTICAL  MECHANICS 

external  and  those  which  are  internal  to  the  system.  Let 
F/  denote  the  resultant  of  the  external  forces  acting  on  the 
particle,  and  F."  denote  the  resultant  of  the  internal  forces 
acting  upon  it,  due  to  its  connection  with  the  rest  of  the 
system.     Then 

F,=  F/+F,". 

But  since  F  is  the  resultant  of  all  the  external  forces  acting 
upon  all  the  particles  of  the  system  we  have 

f  =  n:f/ 

=2Fi-:SF/'. 

The  second  sum  of  the  left-hand  member  is  the  sum  of  the 
internal  forces  and  is  nil,  because  the  internal  forces  come 
in  pairs  which  mutually  annul  each  other.     Therefore 

F  =  2:Fi  (IV) 

=  Smv  (IV) 

=  |(Smv).  (V) 

These  are  results  which  are  worth  noting.  Equation  (IV) 
states  that  the  resultant  external  force  acting  upon  a  system 
equals  and  is  opposite  to  the  vector  sum  (or  the  resultant)  of 
the  kinetic  reactions  of  all  the  particles  of  the  system. 

Equation  (V)  states  that  the  resultant  external  force  acting 
a  pun  a  system  equals  the  time  rate  of  change  of  the  resultant 
momentum  of  the  system. 

PROBLEMS. 

(I)  Show  that  the  component,  along  any  direction,  of  the  resultant 

force  acting  upon  a  particl [uals  the  rate  at  which  the  corresponding 

component  of  its  momentum  changes,  that  is, 

A'  =  —  (mx) ,  etc. 

3how  that   the  component,  along  any  direction,  of  the  resultant 
:  force  acting  upon  a  ByBtem  equals  the  rate  at  which  the  corre- 


IMPULSE  AND  MOMENTUM  J  11 

sponding  component  of  the  resultant  momentum  of  the  system  changes, 
that  is, 

X  =  (jt  (Smi),  etc. 

at 

194.  The  Principle  of  the  Conservation  of  Momentum.  — 
When  the  resultant  external  force  is  zero  equation  (V)  gives 

or  £  (mv)  =  const.  (VI) 

Therefore  when  the  sum  of  the  external  forces  acting  upon  a 
system  vanishes  the  resultant  momentum  of  the  system  remains 
constant,  both  in  direction  and  magnitude.  This  is  the  prin- 
ciple of  the  conservation  of  momentum.  The  momenta  of 
the  various  parts  of  an  isolated  system  may  and,  in  general, 
do  change,  but  the  vector  sum  of  the  momenta  of  all  the 
particles  of  the  system  cannot  change  either  in  direction  or 
in  magnitude. 

PROBLEM. 

Show  that  if  the  component,  along  any  direction,  of  the  resultant 
external  force  vanishes,  the  corresponding  component  of  the  resultant 
momentum  of  the  system  remains  constant,  that  is, 

-///./•  =  const.,  when  X  =  0. 

195.  Momentum  of  a  System.  —  The  magnitude  of  the 
.r-component  of  the  resultant  momentum  of  a  system  may 
be  put  in  the  following  forms: 

2  ///./•=    ( (Smx) 

=  -(ilfi)  [by  equation  (I'),  p.  141] 


=  Mi. 
Similarly  2my  =  My 

Zmz=  Mi,. 


vin 


242  ANALYTICAL  MECHANICS 

where  .1/  is  the  total  mass  and  x,  y,  and  z  are  the  coordi- 
nates of  the  center  of  mass  of  the  system.  Combining  the 
last  three  (Miuations  in  a  single  vector  equation  we  obtain 

2mv  =  Mv,  (VII) 

which  states  that  the  resultant  momentum  of  a  system  equals 
tin  proiluct  of  the  total  mass  of  the  system  by  the  velocity  of  its 
a  nli  r  of  mass. 

196.  Motion  of  the  Center  of  Mass  of  a  System.  —  Com- 
bining equations  (V)  and  (VII)  we  get 

F  =  Mv,  (VIII) 

which  states  that  the  resultant  external  force  acting  upon  a 
system  equals  the  product  of  the  total  mass  of  the  system  by 
the  acceleration  of  its  center  of  mass.  But  equation  (VIII) 
is  the  force  equation  for  a  particle  of  mass  M,  which  is  acted 
upon  by  a  force  F.  Therefore  the  center  of  mass  of  a  system 
moves  as  if  the  entire  mass  of  the  system  were  concentrated  at 
that  point  and  nil  I  he  forces  acting  upon  the  system  were  ap- 
plied to  the  resulting  particle. 

PROBLEM. 

Show  thai  when  the  component,  along  any  direction,  of  the  resultant 
force  acting  upon  a  system  vanishes  the  corresponding  component  of  the 
Velocity  of  the  center  of  mass  remains  constant,  that  is, 

x  =  const.,   when  X  =  0. 

ILLUSTRATIVE   PROBLEM. 

A  bullet  penetrates  a  fixed  plate  to  a  depth  d.  How  far  would  it 
penetrate  if  the  plate  were  free  to  move  in  the  direction  of  motion  of  the 
bullet? 

Let  F  be  the  mean  resisting  force  which  the  plate  offers  to  the  motion 
of  the  bullet.  When  the  plate  is  fixed  all  the  energy  of  the  bullet  is  ex- 
pended  in  doing  work  against  this  force.     Therefore  we  have 

Fd  =  i  mv\  (1) 

where  m  is  the  mass  and  v  the  velocity  of  the  bullet.    When  the  target  is 
•  move  part  of  the  energy  of  the  bullet  is  expended  in  giving  the 


IMPULSK  AND  MOMENTUM  243 

target  and  the  bullet  a  common  velocity  v'.     Therefore  if  d'  be  the  new 
depth  of  penetration  we  have 

Fd'=imvi-i(m  +  M)v,i,  (2) 

where  M  is  the  mass  of  the  target.     Eliminating  F  between  equations 
(1)  and  (2)  we  get 


'-I)-8?©]* 


(3) 


But  by  the  conservation  of  momentum  we  have 

mv=  (m  +  U)v'.  (4) 

Therefore  eliminating  the  velocities  between  equations  (3)  and  (4)  we  get 

d'=--^-J.  (5) 

M  +  in 

It  is  evident  from  equation  (5)  that  when  the  target  is  free,  but  very  large 
compared  with  the  bullet,  the  depth  penetrated  is  about  the  same  as  when 
it  is  fixed. 

PROBLEMS. 

1.  A  particle  which  weighs  2  ounces  describes  a  circle  of  1.5  feet  radius 
on  a  smooth  horizontal  table.  If  it  makes  one  complete  revolution  in 
every  3  seconds  find  the  magnitude  and  direction  of  the  impulse  imparted 
by  the  force,  which  keeps  the  particle  in  the  circle, 

(a)  in  one-quarter  of  a  revolution; 

(b)  in  one-half  of  a  revolution; 

(c)  in  three-quarters  of  a  revolution ; 

(d)  in  one  complete  revolution. 

2.  Find  the  expression  for  the  impulse  imparted  to  a  particle  in  de- 
scribing an  arc  of  a  circle  with  uniform  speed. 

3.  Considering  the  rate  of  change  of  the  momentum  of  a  particle  which 
describes  a  uniform  circular  motion  derive  the  expression  for  the  central 
force. 

4.  If  we  neglect  the  resistance  of  the  air  to  the  motion  of  a  projectile 
what  can  we  state  with  regard  to  the  components  of  the  momentum  in 
the  horizontal  and  vertical  direction-'.' 

6.  A  train  which  weighs  100  tons  runs  due  south  at  the  rate  of  one 
mile  :i  minute.  Find  the  lateral  force  on  the  western  rails  due  to  the 
rotation  of  the  earth,  while  the  train  passes  the  line  of  30°  latitude. 

6.  At  what  latitude  will  the  force  of  the  preceding  problem  be  a  maxi- 
mum?   Determine  its  amount. 


244  ANALYTICAL  MECHANICS 

7.  Two  trains,  weighing  150  tons  each  and  moving  towards  each 
other  at  the  rale  of  40  miles  an  hour,  collide.  Find  the  average  force 
which  comes  into  play  if  the  collision  lasts  1.5  seconds. 

8.  A  body  explodes  while  at  rest  and  flies  to  pieces.  If  at  any  instant 
after  the  explosion  the  different  parts  of  the  body  are  suddenly  connected, 
will  it  move0 

9.  A  shell  of  mass  m  explodes  at  the  highest  point  of  its  flight  and 
breaks  into  two  parts,  the  one  n  times  the  other.  Find  the  velocity  of 
one  piece  if  the  other  is  brought  to  rest  for  an  instant  by  the  explosion. 
The  velocity  of  the  shell  at  the  instant  of  explosion  is  v. 

10.  In  the  preceding  problem  will  the  motion  of  the  center  of  mass  of 
the  entire  shell  be  affected  by  the  explosion?  Answer  this  question  on 
the  assumption  (a)  that  there  is  no  air  resistance,  (b)  that  there  is  an  air 
resistance. 

11.  A  man  walks  from  one  end  to  the  other  of  a  plank  placed  on  a 
smooth  horizontal  plane.     Show  that  the  plank  is  displaced  a  distance 

M     , 


M+m 


where  M  and  m  are  the  masses  of  the  man  and  of  the  plank,  respectively, 
and  I  is  the  length  of  the  plank. 

12.  A  shell,  which  weighs  150  pounds,  strikes  an  armor  plate  with  a 
velocity  of  2000  feet  per  second  and  emerges  on  the  other  side  with  a 
velocity  of  500  feet  per  second.  Supposing  the  resisting  force  to  be  uni- 
form, find  its  magnitude  and  show  that  the  impulse  produced  by  it  equals 
the  change  in  the  momentum  of  the  shell  while  plowing  through  the  plate. 
The  plate  is  10  inches  thick. 

COLLISION  AND  IMPACT. 

197.  Central  Collision.  — -  If  two  bodies  collide  while  moving 
along  the  line  which  joins  their  centers  of  mass  the  collision 
ie  said  to  be  central.  In  order  to  fix  our  ideas  suppose  the 
colliding  bodies  to  be  spheres,  then  Fig.  118  represents 
roughly  the  state  of  affairs  during  the  collision.  For  a  short 
interval  of  time  after  the  spheres  come  into  contact  their 
centers  approach  each  other  and  a  little  deformation  takes 
place  in  the  neighborhood  of  the  point  of  contact  at  the  end 
of  which  the  centers  of  the  spheres  are,  just  for  an  instant, 
with  respect  to  one  another,  and  are  moving  with  a 


IMPULSE  AND  MOMENTUM 


245 


common  velocity.  Then  the  deformed  parts  of  the  spheres 
begin  to  regain,  at  least  partially,  their  original  forms  and 
cause  the  spheres  to  separate. 

The  process  of  collision  may, 
therefore,  be  divided  into  two 
parts.  The  first  part  lasts  from 
the  initial  contact  at  t=  0  until 
the  instant  when  the  centers  of 
the  spheres  are  nearest  together 
at  t=  ti.  The  second  part  be- 
gins at  t  =  U  and  lasts  until  the 
spheres  separate  at  t  =  t\.  The 
impulse  imparted  to  each  body 
during  the  first  part  of  the 
collision  is  called  the  impulse 
of  compression,  while  that  im- 
parted during  the  second  part  is 
called  the  impulse  of  restitution. 

Let  mi  and  m2  be  the  masses 
of  the  colliding  bodies,  vi  and 

v2  be  their  velocities  just  before  and  v/  and  v2'  just  after 
the  collision,  and  let  v  be  their  common  velocity  at  the 
instant  of  maximum  compression,  that  is,  when  the  distance 
between  the  centers  of  mass  is  shortest.  Further,  let  L  and 
L'  denote  the  impulses  of  compression  and  of  restitution, 
respectively.     Then  we  have 

—  m2  (v—  ft), 


Fig.  118. 


L=  fUFdt=  m,  (»- 
Jo 

L'=  ftlFdt=  mi  (ft' 


ft)- 

-v) 


rm  (>.,'  —  v). 


The  foregoing  relations  follow  at  once  from  the  definition 
of  impulse  and  from  the  fad  thai  the  colliding  bodies  form 
a  system  which  is  not  acted  upon  by  external  forces,  and 
consequently  the  sum  of  their  momenta  remains  constant 

during  the  collision. 


246  ANALYTICAL  MECHANICS 

198.  Coefficient  of  Restitution.  —  It  is  found  by  experiment 
that  the  ratio  of  the  impulse  of  restitution  to  the  impulse 
of  compression  depends  only  upon  the  nature  of  the  bodies 
in  collision.  The  ratio,  therefore,  is  a  constant  of  the  sub- 
stances in  collision.  This  constant  is  called  the  coefficient 
of  restitution,  and  is  generally  denoted  by  the  letter  e.     Thus 

•— t  (ix) 


V  —  V\ 

ih'  -  V 


v—  v2 
Eliminating  v  we  obtain 

r'—v' 

e  = •  (X  ) 

Vi  -  V-2 

But  (vi  —  v2)  and  ( -  i\'  +  v%)  are  the  velocities  of  the  first 
body  relative  to  the  second,  just  before  and  just  after  the 
collision.  Denoting  them  by  V  and  V,  respectively,  we 
obtain 

-?  CD 

_  relative  velocity  after  impact 
relative  velocity  before  impact 

199.  Resiliency.  —  When  two  bodies  rebound  after  col- 
lision they  are  said  to  have  resiliency,  and  the  contact 
Le  called  clastic  contact.  The  coefficient  of  restitution  is 
a  i Measure  of  the  resiliency  of  the  colliding  bodies.  When 
e=  1  the  resiliency  of  the  colliding  bodies  is  perfect  and  the 
contact  is  Bald  to  be  perfectly  elastic. 

The  coefficient  of  restitution  cannot  have  a  value  greater 
than  unity,  as  will  be  seen  from  a  consideration  of  the  trans- 
formation of  energy  which  takes  place  during  collision.  At 
the  beginning  of  the  collision  the  bodies  have  a  certain 
amount  of  kinetic  energy  which  depends  upon  their  relative 


IMIMLSi;  AM)   MOMENTUM  JIT 

velocity  at  that  instant.  During  the  compression  pari  of 
the  collision  a  fraction  of  their  energy  is  transformed  into 
potential  energy  of  compression  and  the  rest  into  heat  energy. 
During  the  restitution  a  fraction  of  the  potential  energy  is 
transformed  into  kinetic  energy  and  the  rest  into  heat  energy. 
Thus,  in  general,  the  kinetic  energy  at  the  end  of  the  collision 
is  less  than  that  at  the  beginning.  Therefore  the  relative 
velocity  at  the  end  of  the  collision  is  less  than  thai  at  the 
beginning.  Thus  the  coefficient  of  restitution  is,  in  general, 
less  than  unity.  If  none  of  the  energy,  which  is  due  to  the 
relative  motion  of  the  colliding  bodies,  is  lost  in  the  form  of 
heat,  it  is  all  transformed  into  potential  energy  during  the 
compression  and  back  into  kinetic  energy  during  the  resti- 
tution. In  this  case  the  relative  velocity  at  the  end  of  the 
collision  equals  that  at  the  beginning,  which  makes  the 
coefficient  of  restitution  unity.*  The  relative  velocity  at 
the  end  of  the  collision  may  be  made  greater  than  that 
at  the  beginning  by  having  explosives  at  the  point  of  con- 
tact. But  this  does  not  come  in  the  definition  of  the  coeffi- 
cient of  restitution.  Therefore  unity  is  the  highest  value 
of  e.  When  all  the  kinetic  energy  is  transformed  into  heal 
during  the  collision  the  bodies  have  no  relative  velocity  after 
the  collision.  In  this  case  the  contact  is  called  perfectly 
inelastic.  Evidently  e  is  zero  when  the  contact  is  perfectly 
inelastic.  Therefore  the  value  of  e  lies  between  zero  and 
unity.  The  values  of  the  coefficient  of  restitution  are  0.95 
for  glass  on  glass,  0.81  for  ivory  on  ivory,  and  0.15  for  lead 
on  lead. 

200.  Loss  of  Kinetic  Energy  of  Colliding  Bodies.  -  The 
kinetic  energy  of  a  system  equals  the  kinetic  energy  due 
to  the  linear  motion  of  the  system  with  the  velocity  of  its 

*  In  working  out  problems  in  which  the  contact  is  perfectly  elastic  instead 
of  introducing  the  coefficient  of  restitution  make  use  of  the  principle  «>f  the 
conservation  of  energy.  The  conservation  of  dynamical  energy  holds  only 
when  the  contacl  is  perfectly  elastic.  But  the  conservation  of  momentum 
and  the  conservation  (general)  of  energy  arc  true  under  all  circumstai 


248  ANALYTICAL  MECHANICS 

contor  of  mass  plus  the  kinetic  energy  of  its  parts  due  to  their 
motion  relative  to  the  center  of  mass.  Collision  does  not 
affect  the  mot  ion  of  the  center  of  mass  of  the  system  formed 
of  the  colliding  bodies,  because  the  forces  which  arise  during 
the  collision  are  internal  forces.  Therefore  that  part  of  its 
kinetic  energy  which  is  due  to  the  motion  of  its  center  of 
mass  'Iocs  not  suffer  any  loss.  The  loss  occurs  in  that  part 
of  the  energy  which  is  due  to  the  motion  of  the  parts  of  the 
system  with  respect  to  the  center  of  mass.  Referring  all 
the  velocities  to  the  center  of  mass  and  denoting  the  loss 
of  kinetic  energy  by  Th  we  have 

Tt  =  {\  rriivr  +  \  m2v2-)  -  (§  m^"2  +  \  m2v22), 

where  i\  and  r2  are  the  velocities  just  before  and  vx'  and  v2 
the  velocities  just  after  the  collision. 

We  can  eliminate  Vi  and  v2  from  this  expression  for  Tt  by 
means  of  the  principle  of  the  conservation  of  momentum  and 
the  definition  of  e.     According  to  the  former 

mii\  +  m»v2  =  miVi  +  m2vof 
and  by  (X') 

Vi  —  v2  =  —  e  (i\  —  v2). 

Eliminating  v,_'  between  the  last  two  equations  we  have 

th  -  i\'  =      -^ —  (i\  -  r2)(i  +  e). 
mi  +  m2 

The  following  changes  in  the  expression  of  Tt  are  effected 
by  mean-  of  the  last  three  equations. 

T,  =  Jw.O'i2  -  *'/2)  +  h  ™*  (v22  -  v,'-) 

=  I  »'i  (V,  -  Vi)(Vi  +  iO  -f  J  »h  0>  -  v2')(v2  +  v2) 

=  2  '"I  (''I  -  >'i')0i  -  th  +  Vi    -  V2) 

=  l"iiO\  -  yiOCVi-  ''i)(l  -e) 

\     "^(Vl-v2)2(l-e2).  (XI) 

I'll  +  llh 

When  the  colliding  bodies  are  perfectly  elastic  then  e=l 


[MPULSE  AM)   MOMENTUM  249 

and   Tt  =  0;  on  the  other  hand  if  the  bodies  are  perfectly 
inelastic,  c  =  0;  therefore,  Tt  =  h         -?—  (i'{  —  /•■>)-. 

201.  Impact. — When  the  mass  of  one  of  the  colliding 
bodies  is  very  large  compared  with  that  of  the  other  the 
velocity  of  the  former  with  respect  to  the  center  of  mass  of 
the  colliding  system  does  not  change  appreciably  during  the 
collision.  In  such  a  case  the  body  with  the  greater  mass  is 
considered  to  be  fixed  and  the  collision  is  called  an  impact. 
The  impact  of  a  falling  body  when  it  strikes  the  ground  is 
a  case  in  point. 

The  velocities  of  the  larger  mass  before  and  after  the 
collision,  as  well  as  the  common  velocity  at  the  instant  of 
maximum  compression,  are  negligible.  Therefore  making 
these  changes  in  the  expressions  for  L,  V ',  e,  and  Tt  and 
dropping  the  subscripts  we  obtain 
L  =  mv, 
V  =  —  mv, 

e=v~,  (X") 

v 

and  T,=  \mv-{\-e-),  (XI') 

where  m  is  the  mass  of  the  impinging  body,  while  v  and  v'  are 
its  velocities  just  before  and  just  after  impact,  respectively. 

ILLUSTRATIVE    EXAMPLE. 

A  ball  which  is  thrown  vertically  down  from  a  height  //  rises  to  the  point 
of  projection  after  impinging  against  a  horizontal  floor.  Find  the  ve- 
locity of  projection  and  the  loss  in  energy. 

Let  v0  be  the  velocity  of  projection,  then  the  velocities  just  before  and 
just  after  the  impact  arc 

v  =  V?'o2  +  2  gh     and     v'  =  s/2gh, 

respectively.     But  v'  =  ev.  ^Therefore 

Ti  =  \mv-(l  -c-)=  J  nii'o- 


250  ANALYTICAL   MECHANICS 

Discussion.  The  energy  lost  during  the  impact  equals  the  kinetic 
energy  of  projection,  us  would  be  expected  from  the  conservation  of 
energy. 

When  c  =  1,  v0  =  0  and  Tt  =  0.  In  other  words  when  the  ball  is  per- 
fectly elastic  it  will  rise  to  the  heighl  from  which  it  is  dropped.  The  entire 
kinetic  energy  is  transformed,  during  the  impact,  into  potential  energy  and 
back  to  kinetic  energy  without  any  loss. 

When  e  =  0,  v0  =  ao  and  Tt  =  oo,  that  is,  if  the  contact  is  perfectly 
inelastic  no  value  of  the  velocity  of  projection  will  enable  the  ball  to 
rebound  after  the  impact. 

PROBLEMS. 

1.  Show  that  when  two  perfectly  elastic  spheres  of  equal  mass  collide 
centrally  they  exchange  velocities. 

2.  A  ball  of  mass  mi,  impinging  directly  on  another  ball  of  mass  m2  at 
n  -t ,  comes  to  rest.    Show  that  mi  =  em2. 

3.  Two  perfectly  elastic  balls  collide  directly  with  equal  velocities. 
The  relation  between  their  masses  is  such  that  one  of  them  is  reduced  to 
rest.     Find  the  relation. 

4.  A  ball  which  is  dropped  on  a  horizontal  floor  from  a  height  h  reaches 
a  height  equal  to  $  h  at  the  second  rebound.  Find  the  coefficient  of 
restitution. 

6.   A  metal  patched  bullet  strikes  a  wall  normally  with  a  velocity  of 

1200  — -.     With  what  velocity  will  it  rebound  if  e  =  0.4? 
sec. 

1  +  e 

6.  Show  that  if  two  equal  halls  collide  centrally  with  velocities  v 

and  —  v,  the  one  which  has  the  former  velocity  will  come  to  rest. 

7.  A  bullet  strikes  a  vertical  target  normally  and  rebounds.  Find- the 
relation  between  the  distances  of  the  foot  of  the  target  from  the  rifle  and 
from  the  place  where  the  bullet  strikes  the  ground. 

8.  Two  perfectly  elastic  equal  balls  collide  with  velocities  inversely  as 
their  masses.      Find  the  velocities  after  collision. 

9.  Two  billiard  balls  collide  centrally  with  velocities  of  8  feet  per 
.second  and  16  feel  per  second.     Supposing  e  =  0.8,  find  the  final  velocities. 

10.  A  ball  ia  dropped  from  I  lie  top  of  a  tower,  at  the  same  instant  that 
another  hall  of  equal  ma--  i-  projected  upward  from  the  base  of  the  tower, 
with  a  velocity  jusl  enough  to  raise  it  to  the  top  of  the  tower.  Show  that 
if  the  hall-  collide  centrally  the  falling  hall  will  rise,  on  the  rebound,  to  a 

:  (3  +  e2)  above  the  ground,  where  h  is  the  height  of  the  tower. 

l 


IMPUISE  AM)  MOMENTUM  251 

11.  Two  spheres  of  masses  111  and  2  m  moving  with  equal  velocities 

alon^  two  lines  at  righl  angles  to  each  other  collide  at  the  instant  when 
their  .enters  are  on  the  line  of  motion  of  the  smaller  sphere.  Show  that  it' 
t  he  contact  is  smooth  and  e  =  0.5  the  smaller  sphere  will  come  to  rest ,  and 
find  the  direction  and  magnitude  of  the  velocity  of  the  larger  sphere. 

12.  In  the  preceding  problem  let  m  =  500  gm,  v  =  40  cm.  per  second 
and  find 

(a)  the  impulse,  its  magnitude  and  direction; 

(b)  the  loss  of  energy. 

13.  A  metal  patched  bullet  which  weighs  1.5  ounces  strikes  a  rock, 
normally,  with  a  velocity  of  1500  feet  per  second.  Find  the  velocity  with 
which  it  will  rebound  and  the  impulse  given  to  the  rock;   e  =  0.5. 

14.  A  body  impinges  against  another  body  which  has  a  mass  n  times 
as  large.     Show  that  if  the  larger  body  is  at  rest  and  the  contact  inelasl  ic 

the  loss  of  energy  is  — — —  times  its  value  before  the  collision. 
n  -f-  1 

15.  A  particle  is  projected  up  a  smooth  inclined  plane  with  a  velocity 
y/gh;  simultaneously  a  particle  of  equal  mass  is  allowed  to  slide  down 
the  inclined  plane.  The  two  collide  somewhere  on  the  plane.  Find  the 
velocities  with  which  the  particles  will  arrive  at  the  bottom  of  the  plane. 
h  =  the  height  of  the  inclined  plane. 

16.  Two  small  spheres  of  masses  m  and  2  m  move  in  a  smooth  circular 
groove  on  a  horizontal  table  with  equal  speeds  in  opposite  directions. 
Find  the  position  of  the  second  collision  relative  to  the  first;  e  =  0.6. 

17.  In  the  preceding  problem  find  the  interval  of  time  between  the  first 
and  the  eleventh  collision,  under  the  following  assumptions  —  the  radii 
of  the  particles  are  negligible  compared  with  that  of  the  circular  groove, 
which  equals  50  cm.,  the  common  speed  of  the  spheres  just  before  the  first 
collision  is  500  cm.  per  second,  the  time  of  collision  is  negligible. 


202.  Efficiency  of  a  Blow.  —  A  blow  may  be  struck  to  pro- 
duce one  or  the  other  of  two  distinct  results.  The  object  of 
a  blow  from  a  hammer  in  driving  a  nail  is  quite  different  from 
that  of  a  blow  in  shaping  a  rivet.  Efficiency  in  the  first 
case  means  gn  at  esl  amounl  of  driving  with  the  least  amount 
of  deformation,  while  in  the  second  case  it  means  greatest 
amounl  of  smashing  with  the  least  amounl  of  driving. 
Therefore  the  efficiency  of  a  blow  is  different  for  these  two 
cases.     We  may  define 


252  ANALYTICAL  MECHANICS 

~  .   .         „  .  energy  expended  in  driving 

Driving  efficiency  = 7; — s—  — -,    ,      • 

total  energy  expended 

0        ,  .         ~  .  energy  expended  in  deforming 

Smashing  efficiency  = ~\  .  v 3—3 — &- 

total  energy  expended 

Consider  the  ease  of  a  blow  which  drives  a  nail  or  a  pile.  Let 
.1/  be  the  mass  of  the  hammer,  m  the  mass  of  the  pile,  v  the 
velocity  of  the  hammer  just  before  impact,  v'  the  velocity 
just  after  impact.  The  contact  between  the  hammer  and 
the  pile  may  be  regarded  as  inelastic,  therefore  just  after  the 
impact  both  the  pile  and  the  hammer  have  the  same  velocity 
1/.  In  other  words,  immediately  after  the  impact  there  is  an 
amount  of  energy  equal  to  \  {M  +  m)  v'2  available  for  driv- 
ing the  pile,  while  the  balance  of  the  energy  of  the  ham- 
mer, that  is,  \  Mv2  —  \  (M  +  m)  v'2,  is  expended  during 
the  impact  in  producing  permanent  deformation,  heat,  and 
sound.  Substituting  these  in  the  two  definitions  for  the 
efficiency  of  a  blow  we  obtain 

(M+m)  v'°- 


Driving  efficiency 
Smashing  efficiency  =  1 


Mv2 

(M  +  m)  v'2 
Mv2 


Immediately  after  the  impact  practically  all  the  momentum 
of  the  hammer  relative  to  the  earth  will  be  in  the  hammer 
and  the  pile;  therefore  we  can  write 
Mv—  {M+m)  v'. 

Eliminating  the  velocities  between  the  last  equation  and  the 
above  expressions  for  the  two  efficiencies  we  obtain 

M 


Driving  efficiency  = 


M+m 


Smashing  efficiency 


(XII) 


M+m 

It  Is  evident  from  these  expressions  that  for  driving  piles  or 
oailfl  the  ram  or  the  hammer  head  must  have  a  large  mass 


IMPULSE  AND  MOMENTUM  253 

compared  with  the  pile  or  the  nail.  On  the  other  hand,  for 
shaping  rivets  the  anvil  must  have  a  large  mass  compared 
with  the  hammer. 

203.  Motion  where  Moving  Mass  Varies.  —  If  the  moving 
mass  varies,  as  in  the  case  of  an  avalanche,  the  relation  be- 
tween impulse  and  momentum  still  holds,  that  is,  impulse 
equals  the  change  in  the  momentum.  We  have  to  take  into 
account,  however,  the  change  in  the  momentum  of  the  mass 
which  is  continually  added  to  the  moving  system  as  well  as 
that  of  the  original  mass.  Let  a  mass  dm  be  added  in  the 
time  dt;  then  supposing  dm  to  have  been  initially  at  rest  the 
total  change  in  the  momentum  is  m  dv  +  v  dm,  where  the  first 
term  is  the  increase  in  the  momentum  of  m  and  the  second 
term  is  the  increase  in  the  momentum  of  dm.  Therefore  the 
impulse  given  by  the  resultant  force  dF  in  the  time  dt  is 

F  dt  =  m  c/v  +  v  dm, 

_         dv   .     dm 

which  is  the  same  equation  as  (III),  except  that  in  (III)  m 
was  considered  to  be  constant,  while  here  it  is  considered  as 
a  variable. 

If  dm  has  an  initial  velocity  u,  then  the  change  in  the 
momentum  of  dm  is  (v  —  u)  dm.     Therefore 

_  dv    .    /  dm 

F=,„-+(v-u1(// 

d  f      \        dm  x  , , , 


254 


ANALYTICAL  MECHANICS 


ILLUSTRATIVE  EXAMPLES. 

1.  A  jet  of  water  strikes  a  concave  vessel  with  a  velocity  of  80  feet  per 
second  and  then  leaves  it  with  a  velocity  which  has  the  same  magnitude 
as  the  vetocity  of  impact  but  makes  an  angle  of  120°  with  it.  If  the 
diameter  of  the  jel  is  I  inch  find  the  force  necessary  to  hold  the  concave 
vessel  in  position. 

The  force  experienced  by  the  vessel  equals  the  rate  at  which  it  receives 
moment uin.  Suppose  the  vessel  to  be  symmetrical  with  respect  to  the 
axis  of  the  jet,  as  in  Fig.  119,  then  by  symmetry  there  can  be  no  resultant 
force  on  the  vessel  in  a  direction  perpendicular  to  the  axis  of  the  jet. 
Therefore  we  need  to  consider  only  the  change  in  momentum  along  the 
axis.  Let  m  be  the  mass  of  water  delivered  by  the  jet  in  the  time  t, 
v  the  velocity  of  impact,  and  a  the  change  in  the  direction  of  flow.  Then 
the  force  is  a  V 

—  mv  cos  a  i — - 


Fig.  119. 


where  .1  is  the  area  of  the  cross-section  of  the  jet  and  wi  is  the  weight  of 
:i  cubic  foot  of  water.  Replacing  the  various  magnitudes  by  their  nu- 
merical  values  we  obtain 


■»g*'(s<0'*-(w=:)'*(' 


+  cos  60°) 


32 


■'  L02.3  lb. 


I  Mm  i  S8ION.        It   is  evident   from  the  general  expression  of  F  that  its 
value  depends  upon  a  and  varies  between  zero  for  a  =  0  and  "  "'"  ''  for 


Whe 


'J' 


IMPULSE  AXD  MOMENTUM  255 

2.  A  uniform  chain  is  hung  from  its  upper  end  so  thai  its  lower  end 
just  touches  an  inelastic  horizontal  table,  and  then  it  is  allowed  to  fall. 
Find  the  force  which  the  table  will  experience  at  any  instanl  during  the 
fall  of  the  chain. 

The  force  is  partly  due  to  the  weight  of  that  part  of  the  chain  which  is 
on  the  table  at  the  instant  considered  and  partly  due  to  the  rate  at  which 
the  table  is  receiving  momentum.  Let  x  be  the  height  of  the  upper  end 
of  the  chain  above  the  table,  /  the  total  length,  and  p  the  mass  per  unit 
length.  Then  pg  {I  -  x)  is  the  weight  of  that  part  of  the  chain  which  is 
on  the  table.  On  the  other  hand  the  momentum  which  the  table  receives 
in  the  interval  of  time  dt  is  pv  <lt  •  v.  Therefore  the  rate  at  which  it  re- 
ceives momentum  is  pr'2,  where  v  is  the  velocity  of  that  part  of  the  chain 
which  is  above  the  ground.  This  velocity  is  the  same  as  that  of  the  upper 
end  of  the  chain,  therefore 

v  =  V2g(l-x). 

Hence  the  total  force  is 

F  =  p(l-x)g  +  p-2gQ-x) 

=  3p(l-x)g. 

Discussion.  —  When  x  =  I,  that  is,  at  the  beginning  of  the  motion, 
the  force  is  zero.  When  x  =  0,  that  is,  at  the  end  of  the  motion,  it  is 
3  pig,  or  three  times  the  weight  of  the  chain.  As  soon  as  the  entire 
chain  comes  to  rest  on  the  table  the  force  equals  the  weight  of  the 
chain. 

3.  A  spherical  raindrop,  descending  by  virtue  of  its  weight,  receives 
continuously,  by  precipitation  of  vapor,  an  accession  of  mass  proportional 
to  the  surface.     Find  the  velocity  at  any  instant. 

The  external  force  acting  upon  the  drop  at  any  instant  equals  the  rate 
at  which  its  momentum  changes,  therefore 

mg  =  'jt  (m»),  (1) 

where  m,  the  mass  of  the  drop,  is  variable.  Since  the  accession  of  mass  is 
proportional  to  the  surface  the  rate  of  change  of  radius  of  the  drop  will  be 
constant,  bet  <i  he  the  radius  of  the  drop  when  it  begins  to  fall,  r  its 
radius  at  any  later  instant,  and  /.•  the  rate  at  which  r  increases.  Then  at 
any  instant 

m  =  r 

=  T  *7T  (O  +  H)\ 


256  ANALYTICAL  MECHANICS 

where  r  is  the  density  of  water.     Substituting  this  expression  for  m  in 
equation  (1) 

?  (a +  &)»=!  [(a +  &)»»] 

=  (a  +  kt)3~  +  3(a-t-kty-kv 
at 

■'""'  j  +  «-?h'  =  s'  (2' 

the  integral  of  which  is  t;  =  e     a+^         JG/e  a  +  *'    eft  +  c    • 

,..       r  =  e-3log  (a  +  *fl  I"  J^log  (a  +  fcQ  ^  _j_  cl 

=  (a  +  kt)->[gf(a  +  kt)*dt  +  c] 

=  (a  +  A-0~3Q  (4  a3<  +  6  a-kt2  +  4  akH3  +  *#*)  +  c] . 

Let     v  =  0  when  <  =  0;     then    c  =  0. 
.  g<  4  a3  +  6  a2A-/.  +  4  aF-<2  +  fc3<3 

4  (a  +  fa)» 


?('  +  ;+?+?)• 


PROBLEMS. 

1.  Find  the  pressure  upon  the  canvas  roof  of  a  tent  produced  by  a 
Bhower.     The  following  data  are  given  —  the  raindrops  have  a  velocity  of 

50  —  at  right  angles  to  the  roof;  the  intensity  of  the  shower  is  such  as 

to  produce  a  deposit  of  0.2  inch  per  hour;  1  cubic  foot  of  water  weighs 
62.5  pounds. 

2.  Find  the  pressure  on  horizontal  ground  due  to  the  impact  of  a 
column  of  water  which  falls  vertically  from  a  height  of  500  feet. 

3.  Water  flowing  through  a  pipe  at  the  rate  of  100  - — '  is  brought  to 
*  Equation   (2)  is  of  the  form-^  +Py  =  Q,  which  is  the  typical  linear 

equation,  with  the  integral  y  =  e~fPdxT  fQeJ*** dx  +  c~|. 


IMPUISE  AND  MOMENTUM 


257 


rest  in  0.1  second  by  closing  a  valve  at  the  lower  end.  Find  the  in- 
crease of  pressure  produced  near  the  valve  in  both  the  C.Ci.S.  and  the 
British  units.    The  Length  of  the  pipe  is  500  meters. 

4.  A  jet  of  water  strikes  a  blade  of  a  turbine  normally,     [f  the  velocity 

of  the  jet  is  150  feet  per  second,  find  the  pressure  it  exerts  on  the  blade, 

(a)  when  the  blade  is  fixed;  (b)  when  it  has  a  velocity  of  50  feet  per 
second  along  the  jet. 

6.  Figure  120a  represents  a  horizontal  trough  with  smooth  vertical 
walls.  The  stream  is  supposed  to  have  the  same  speed,  6  miles  per  hour, 
in  all  three  parts  of  the  trough.  The  stream  in  C  is  one-third  of  that  in  B. 
Find  the  force  on  the  wall  BC.  The  cross-section  of  the  stream  in  A  is 
5  feet  by  3  feet. 

6.   In  the  preceding  problem  suppose  the  branch  C  to  be  closed. 


Pig.  120. 

7.  A  stream  of  water  flowing  in  a  horizontal  direction  is  divided  into 
two  equal  streams,  as  shown  in  Fig.  120b.  Supposing  the  velocity  of 
the  water  to  remain  unchanged  derive  an  expression  for  the  force  on  the 
obstacle,  and  discuss  it  for  special  values  of  6. 

8.  In  the  preceding  problem  suppose  the  velocity  of  the  stream  to  lie 
5  miles  per  hour,  its  cross-section  before  it  is  divided  to  be  1  feet  by 
2  feet,  and  0  =  120°. 

9.  In  the  precedin'j;  problem  take  6  =  tt. 

10.  In  (S)  take0  =  '^f- 

11.  In  (S)  takefl  =  2tt. 

12.  A  machine  gun  delivers  500  bullets  per  minute  with  a  velocity  of 
L800  feet  per  second.  If  the  bullets  weigh  <>.:>  ounce  each  find  the  average 
force  on  the  carriage  of  the  trim. 

13.  A  train   scoops  up   1500  pounds  of  water  into   the  tender  from   a 

trough  .">()()  yards  long  while  making  50  miles  per  hour.     Find  the  added 
resistance  to  the  motion  of  the  train. 


258 


ANALYTICAL  MECHANICS 


204.  Oblique  Impact  of  a  Particle  upon  a  Fixed  Plane.  Case 
I.  Smooth  Contact.  —  Let  vt  and  vn,  Fig.  121,  be  the  compo- 
nents of  the  velocity  along  the  plane  and  along  the  normal, 
respect ively,  just  before  the  impact;  and  let  vt'  and  vn'  be 
the  corresponding  components  just  after  the  impact.  Since 
the  plane  is  smooth,  no  hori- 
zontal forces  arise  during  the 
impact;  hence  the  horizontal 
component  of  the  momentum 
remains  constant.  Therefore 
mv,  =  mv,', 


So  far  as  the  vertical  compo- 
nent is  concerned  the  impact 
is  direct;  therefore 


Fig.  121. 


Denoting  by  a  and  0  the  angles  which  the  resultant  velocity 
makes  with  the  normal  just  before  and  just  after  the  im- 
pact we  obtain 


tan  a  =  —  i 

vn 


tan 


.-.     tan  a  =  e  tan  0.  (XIV) 

Discussion.  —  When  the  contact  is  perfectly  clastic  c=  1;  therefore 
the  angle  of  incidence  equals  the  angle  of  reflection  as  in  the  case  of  the 
reflection  of  light.  In  this  case  the  magnitude  of  the  velocity  is  not 
changed  by  the  impact,  as  is  to  be  expected  from  the  conservation  of 
energy.  When  the  contact  is  imperfectly  elastic  the  angle  of  reflection 
I  ween  -  and  the  angle  6f  incidence,  while  the  normal  component  of 

the  velocity  and  consequently  the  magnitude  of  the  total  velocity  is  di- 
minished.    When  the  contact  is  perfectly  inelastic  c  =  0,  and  since  a  is 
•  '  nm-i  be  -   in  order  that  <  tan  0  may  have  a  finite  value.     There- 
fore in  this  case  the  particle  slides  along  the  plane  after  the  collision. 


IMPULSE  AND  MOMENTUM  259 

205.  Case  II.  Rough  Contact.  —  When  the  plane  is  rough 
frictional  forces  come  into  play  and  change  the  tangential 
component  of  the  momentum.  Let  F  be  the  tangential  force 
due  to  friction,  N  the  normal  force,  and  ju  the  coefficient  of 
friction;  then  we  have 


rr 
L„  =  I    Ndt=  —  mvn, 

Ln'=  fT  Ndt  =  mvn'; 

e=-TL»'-v», 

Lt=  I     Fdt=   \    vNdt  = 

Jo                 Jo 

—  nmvn, 

rr               rr 
L/=  /     Fdt=   /     nNdt  = 
Jt               Jt 

—  etxmvn 

But  Lt+  Lt'  =  mvt  —  mvt. 

Therefore  m  (v/  —  vt)  =  —  m/i  (1  +  e)  vn 

and  v/  =  vt  -  n  (I  +  e)  vn. 

Substituting  this  value  of  v/  in  the  expression  for  tan  /3, 
which  is  obtained  from  Fig.  121,  we  get 

Eliminating  vt  between  the  last  equation  and  the  relation 
tan  <x=  —  we  obtain 

e  tan  /3  =  tan  a  —  n  (1  +  e).  (XV) 

Discussion.  —  When  n  =  0,  equation  (XV)  reduces  to  equation  (XIV). 
When/x  =  oo,  tan/3  =  —  oo  or  j3  =  —  -;  therefore  the  particle  slides  along 

the  plane  towards  the  left.    When  e  =  o  and  tan  a  >  n,  tan  /3  =  oo  and 
fi  =   -;  therefore  the  particle  slides  along  the  plane  towards  the  right. 


260  ANALYTICAL  MECHANICS 

When  e  =  0  and  tan  a  <  n,  tan  0=  —  oo  and  0  =  -  ^  ;  therefore  the  particle 
is  reflected  towards  the  left  and  slides  along  the  plane. 

PROBLExMS. 

1.  A  perfectly  elastic  ball  impinges  obliquely  on  another  ball  at  rest. 
Prove  i  bat  their  masses  are  equal  if,  after  impact,  the  balls  move  at  right 
angles. 

2.  A  billiard  ball  strikes  simultaneously  two  billiard  balls  at  rest,  and 
<(iincs  to  rest.     Show  that  the  coefficient  of  restitution  is  f. 

3.  A  particle  slides  down  a  smooth  inclined  plane  and  then  rebounds 
from  a  horizontal  plane.     Find  the  range  of  the  first  rebound. 

4.  A  bullet  strikes  a  target  at  45°  and  rebounds  at  the  same  angle. 

Prove  that  c  =     ~  *,  where  /jl  is  the  coefficient  of  friction. 
1  +M 
6.    Pour  smooth  rods,  which  form  a  square,  are  fixed  on  a  smooth 
horizontal  plane.     A  particle  which  is  projected  from  one  corner  of  the 
square  strikes  an  adjacent  corner  after  three  reflections;  show  that 

e(l  +  e) 


tan  a 


l  +  e(l  +  e) 


where  a  is  the  angle  the  initial  velocity  makes  with  the  rod  joining  the  two 
corners  and  e  is  the  coefficient  of  restitution. 

6.  In  the  preceding  problem  discuss  the  values  of  a  for  special  values 
of  c. 

7.  1  )erive  an  expression  for  the  percentage  of  energy  lost  during  oblique 
impact  (a)  when  the  contact  is  smooth;  (b)  when  the  contact  is  rough. 

8.  Two  billiard  balls  which  are  in  contact  are  struck,  simultaneously, 
by  a  third  ball  moving  with  a  velocity  v,  in  a  direction  perpendicular  to 
the  line  of  cciiters  of  the  first  two.  Supposing  the  table  to  be  perfectly 
smooth  find  the  velocity  of  each  ball  after  impact. 

9.  In  the  preceding  problem  obtain  the  expression  for  the  loss  of 
energy  and  find  its  value  for  the  following  special  cases.  The  balls  weigh 
ti  ounce-  each. 

(a)  V  ■  16  feet  per  second,  c  =  0.8. 
(1>)  v  =  20  feet  per  second,  e  =  0.5. 

10.  A  ball  impinges  againsl  another  ball  which  has  twice  as  large  a 
mass  and  is  at  reel .  The  -mailer  ball  has  a  velocity  of  60  feet  per  second 
i,.  a  direction  which  makes  135°  with  the  line  of  centers.     Find  the  veloci- 

fter  unpad ;  e  -  <>.">. 


IMPULSE  AND  MOMENTUM  261 

GENERAL   PROBLEMS. 

1.  A  gun  is  free  to  move  on  smooth  horizontal  tracks.     Show  that  tin: 

Lobs  of  energy  due  to  recoil  is  -rz E.  where  .1/  and  m  arc  the  ma 

M  +  m 

the  gun  and  the  projectile  respectively,  and  E  is  the  kinetic  energy  which 

is  transmitted  to  the  gun  and  projectile. 

2.  In  the  preceding  problem  compare  the  velocities  of  the  projectile 
when  the  gun  is  fixed  and  when  free  to  move.     Also  show  that  the  actual 

angle  a  at  which  the  projectile  leaves  the  gun  is  given  by  tan  a  =  — — — 

tan  a,  where  a  is  the  angle  which  the  gun  makes  with  the  horizon. 

3.  A  man  stands  on  a  plank  of  mass  m,  which  is  on  a  perfectly  smooth 
horizontal  plane.  He  jumps  upon  another  plank  of  the  same  mass,  then 
back  upon  the  first  plank.  Find  the  ratio  of  the  velocities  of  the  two 
planks  if  the  mass  of  the  man  is  M. 

4.  A  stream  of  water  delivering  1000  gallons  per  minute,  at  a  velocity 

of  20  — -,  strikes  a  plane  (1)  normally,  (2)  at  an  angle  of  30°.     Find  the 
sec. 

force  exerted  on  the  plane. 

5.  A  uniform  chain  is  held  coiled  up  close  to  the  edge  of  a  smooth  table, 
with  one  end  hanging  over  the  edge.  Discuss  the  motion  of  the  chain 
when  it  is  allowed  to  fall,  supposing  the  part  hanging  over  the  edge  to  be 
very  small  at  the  start  of  the  motion. 

6.  In  the  preceding  problem  show  that  the  acceleration  is  constant  if 
the  density  of  the  chain  varies  as  the  distance  from  that  end  of  the  chain 
which  is  in  motion. 

7.  A  mass  of  snow  begins  to  slide  down  a  regular  slope,  accumulating 
more  snow  as  it  moves  along,  thus  forming  an  avalanche.  Supposing  the 
path  cleared  to  be  of  uniform  depth  and  width,  show  that  the  acceleration 
of  the  avalanche  is  constant. 

8.  A  ball  falls  on  a  floor  from  a  height  h  and  rebounds  each  time  verti- 
cally. 

(a)  Show  that  T  =  J-ii  V/^, 

1  —  e  f    g 

where  T  is  the  total  time  taken  by  the  ball  to  come  to  rest.     Find  the 
value  of  T  for  /;  =  25  feet  and  c  =  0.5. 

(b)  Show  that  II  =  \^f2h, 

where  //  is  the  total  distance  described.     Find  the  value  of  //  for  h  =  25 
feet  and  e  =  0.5. 


262  ANALYTICAL  MECHANICS 

9.  A  shell  explodes  at  the  highest  point  of  its  path  and  breaks  up  into 
two  parts,  the  centers  of  mass  of  which  lie  in  the  line  of  motion.  Find 
the  velocity  of  the  pieces  jusl  after  explosion,  taking  m  for  the  mass  of  the 
shell,  n  for  the  ratio  of  the  masses  of  the  pieces,  v  for  the  velocity  of  the 
shell  just  before  explosion,  and  E  for  the  energy  imparted  to  the  pieces 
by  the  explosion. 

10.  A  particle  slides  down  a  smooth  inclined  plane  which  is  itself  free 
to  move  on  a  smooth  horizontal  plane.     Discuss  the  motions  of  the  particle 

and  of  the  plane. 

11.  After  falling  freely  through  a  height  h  a  particle  of  mass  m  begins 
to  pull  ui>  ;l  greater  mass  M,  by  means  of  a  string  which  passes  over  a 
smooth  pulley.     Find  the  distance  through  which  it  will  lift  M. 

12.  A  smooth  inclined  plane  which  is  free  to  move  on  a  smooth  hori- 
zontal plane  is  so  moved  that  a  particle  placed  on  the  inclined  plane 
remains  at  rest.     Discuss  the  motion  of  the  plane. 

13.  A  disk  and  a  hoop  slide  along  a  smooth  horizontal  plane  with  the 
Bame  velocity  v,  then  begin  to  roll  up  the  same  rough  inclined  plane. 
How  high  will  each  rise? 

14.  A  ball,  moving  with  a  velocity  v,  collides  directly  with  a  ball  at  rest. 
sond  ball  in  its  turn  collides  with  a  third  ball  at  rest.     If  the  masses 

of  the  first  and  last  ball  are  ///,  and  m„  respectively,  show  that  the  velocity 
acquired  by  the  third  ball  is  greatest  when  the  mass  of  the  second  ball 
satisfies  the  relation  m2  =  y/m^nt. 

16.  Find  the  maximum  velocity  acquired  by  the  third  ball  of  the  pre- 
ceding problem. 

16.  A  billiard  ball,  moving  at  right  angles  to  a  cushion,  impinges 

directly  on  an  equal  ball  at  rest  at  a  distance  d  from  the  cushion.     Show 

2  i  - 
that  they  will  meet  again  at  a  distance— — d  from  the  cushion. 

1  +  e 

17.  A  ball  is  projected  from  the  middle  point  of  one  side  of  a  billiard 
table,  so  that  it  Btrikesan  adjacent  side  first,  then  the  middle  of  the  opposite 
side.    Show  that  if/  is  the  length  of  the  adjacent  side,  the  ball  strikes  the 

adjacent  side  at  a  point  from  the  corner  it  makes  with  the  opposite 

1  -p  € 

side. 

18.  A  simple  pendulum  hanging  vertically  has  its  bob  in  contact  with 

d  wall.     The  bob  IS  pulled  away  from  the  wall  and  then  it  is  let  go. 
the  coefficient  of  restitution  find  the  time  it  will  take  the  pendulum 
to  come  to  rest. 

19.  A  particle  strikes  a  smooth  horizontal  plane  with  a  velocity  v, 


IMPULSE  AND  MOMENTUM  263 

making  an  angle  a  with  the  plane,  and  rebounds  time  after  time.     Prove 

that 

LVsi„«  B=4^i« 

g{i-e)  g(i-e) 

where  T  is  the  total  time  of  flight  after  the  firsl  impact,  R  the  total 
range,  and  <  the  coefficient  of  restitution. 

20.  In  the  preceding  problem  find  the  values  of  T  and  li  for  the  fol- 
lowing special  eases: 

(a)  v  =  500  meters  per  second,       a  =  30°,       e  =  0.5. 

(b)  v  =  500  meters  per  second,       a  =  90°,       c  =  0.9. 

21.  A  particle  is  projected  horizontally  from  the  top  of  a  smooth 
inclined  plane.  Derive  an  expression  for  the  time  at  the  end  of  which 
the  particle  stops  rebounding  and  slides  down  the  plane.  Compute  its 
value  for  the  following  special  cases: 

(a)  v0  =  500  feet  per  second,     a  =  45°,  c  =  0.5. 

(b)  v0  =  500  feet  per  second,     a  =  30°,  e  =  0.3. 

22.  In  the  preceding  problem  find  the  distance  the  particle  moves 
along  the  plane  before  it  stops  rebounding. 

23.  In  problem  21  find  the  velocity  of  the  particle  at  the  instant  it 
stops  rebounding. 

24.  A  bead  slides  down  a  smooth  circular  wire,  which  is  in  a  vertical 
plane,  and  strikes  a  similar  bead  at  the  lowest  point  of  the  wire.  If  during 
the  collision  the  first  bead  comes  to  rest,  show  that  the  second  bead  will 
rise  to  a  height  e2h  and  on  its  return  will  follow  the  first  head  to  a  heighl 
e*  (1  —  e)2  h,  where  h  is  the  height  from  which  the  first  bead  falls. 

25.  Two  equal  spheres,  which  are  in  contact,  move  in  a  direction  per- 
pendicular to  oieir  line  of  centers  and  impinge  simultaneously  on  a  third 
equal  sphere  which  is  at  rest.  Supposing  the  contacts  to  be  perfectly 
smooth  and  elastic  find  the  velocity  of  each  sphere  after  the  collision. 

26.  A  bullet  hits  and  instantly  kills  a  bird,  while  passing  the  highest 
point  of  its  trajectory.  Supposing  the  bullet  to  stay  imbedded  in  the  bird, 
and  the  bird  to  have  been  at  rest  when  shot,  find  the  distance  between  the 
place  of  firing  and  the  point  where  the  bird  strikes  the  ground. 

27.  Two  particles  of  masses  m,\  and  ///,.  are  connected  by  an  inextensiUe 
string  of  negligible  mass.  The  second  particle  is  placed  on  a  smooth  hori- 
zontal table  while  the  first   is  allowed  to  fall  from  the  e,lure  of  the  table. 

When  the  falling  particle  reaches  a  distance /i  from  the  top  of  the  table  the 

string  becomes  tight.      Find  the  velocity  with  which  the  second  particle 
begins  to  move. 


264  ANALYTICAL  MECHANICS 

28.  A  uniform  chain  Lies  in  a  heap  close  to  the  edge  of  a  horizontal 
table.  One  end  of  the  chain  is  displaced  from  the  edge  of  the  table  so 
that  it  begins  to  fall.     Show  that  when  t  lie  last  portion  of  the  chain  leaves 

the  table  the  chain  will  have  a  velocity  of  \-£->  where  I  is  the  length  of 

tin'  chain. 

29.  A  uniform  plank  is  placed  along  the  steepest  slope  of  a  smooth 
inclined  plane  Show  that  if  a  man  runs  down  the  plank  making  its 
length  in  the  time  given  by 

2  M  a 


t2  = 


M  +  m    g  sin  a 


the  plank  remains  stationary  during  his  motion. 

30.  A  number  of  coins  of  equal  mass  are  placed  in  a  row  on  a  smooth 
horizontal  plane,  each  coin  being  in  contact  with  its  two  neighbors.  A 
similar  coin  is  projected  along  the  line  of  the  coins  with  a  given  velocity. 
Find  the  velocity  with  which  the  last  coin  will  start  to  move. 

31.  A  ball  of  mass  ///,  which  is  at  rest  on  a  smooth  horizontal  plane, 
is  tied  by  means  of  ;i  string  to  a  fixed  point  at  the  same  height  as  the 
center  of  the  ball.  A  second  ball  of  equal  radius  but  of  mass  m'  is  pro- 
jected  along  the  plain'  with  a  velocity  v  which  makes  an  angle  a  with  the 
Btring.  The  second  ball  collides  with  the  first  centrally  and  gives  it  a 
velocity  '/.     Show  that 

in'  sin  a  (1  +  e) 

u  = ; /  ■    ,       v. 

m  +  m  sin-  a 


CHAPTER   XIII. 
ANGULAR   IMPULSE   AND   ANGULAR   MOMENTUM. 

206.  Angular  Impulse.  —  The  mechanical  results  produced 
by  a  torque  may  be  measured  in  two  ways.  If  the  torque 
is  considered  to  act  through  an  angle  the  result  measured 
is  the  work  done  by  the  torque;  on  the  other  hand  if  the 
torque  is  considered  to  act  during  an  interval  of  time  the 
result  measured  is  called  angular  impulse. 

The  angular  impulse  which  a  constant  torque  imparts  to  a 
body  in  an  interval  of  time  equals  the  product  of  the  torque 
by  the  interval  of  time.  If  H  denotes  the  angular  impulse, 
G  the  torque,  and  t  the  time  of  action,  then 

H  =  G  •  t.  (I') 

When  a  vector  is  multiplied  by  a  scalar  the  product  is  a 
vector  which  has  the  same  direction  as  the  original  vector. 
Therefore  H  is  a  vector  and  has  the  same  direction  as  G. 

When  torque  is  not  constant  angular  impulse  equals  the 
vector  sum  of  infinitesimal  impulses  imparted  during  infini- 
tesimal intervals  of  time.     Therefore 


H-f 

Jo 


Gdt 
I'l  (o dt 
=  /    /  tfto, 


(I) 


where  o>0  and  w  are  the  angular  velocities  at  the  beginning 
and  at  the  end  of  the  interval  of  time  during  which  the  torque 
acts. 


acts. 

265 


2G6  ANALYTICAL  MECHANICS 

When  /,  the  moment  of  inertia,  remains  constant,  as  in 
the  case  of  a  rigid  body  rotating  about  a  fixed  axis,  the 
last  integration  can  be  performed  at  once  and  the  following 
result  obtained : 


I 


G  dt  =  Iu  -  /co0.  (II) 


207.  Angular  Momentum.  —  The  magnitude  7o>  is  called 
angular  momentum  and  is  defined  as  the  product  of  the 
moment  of  inertia  by  the  angular  velocity.  Since  I  is  a 
scalar  /a>  is  a  vector  which  has  the  same  direction  as  <o. 
Equation  (II)  states  that  angular  impulse  equals  the  change 
in  the  angular  momentum. 

208.  Moment  of  Momentum. — Angular  momentum  is  often 
called  moment  of  momentum,  because  the  former  may  be 
considered  as  the  moment  of  the  linear  momenta  of  the 
particles  of  the  system  under  consideration.  Let  dm  be  an 
element  of  mass,  r  its  distance  from  the  axis  of  rotation,  and 
r  ite  linear  velocity.  Then  the  moment  of  the  momentum 
of  dm  about  the  axis  is  r-vdm.  Therefore  the  total  moment 
of  momentum  is 


Jf'm  nm 

r • v dm  =  I    r -ru  dm 
u  Jo 

=  /    r2  dm  •  co 


=  /«, 

which  is  the  angular  momentum. 

209.  Dimensions  and  Units.  -Substituting  the  dimensions 
of  G.  i,  I,  and  co  in  equation  (II)  we  find  that  both  angu- 
lar impulse  and  angular  momentum  have  the  dimensions 

M///T  ').      The    units   are   also   the  same   for    both.      The 

3.  unit  is  K  and  the  British  unit  is  ft.  lb.  sec. 


ANGULAB    [MPULSE  AND  ANGULAR   MOMENTUM     2G7 


210.  Torque  and  Angular  Momentum.  —  When  the  moment 
of  inertia  of  a  body  remains  constant  under  the  action  of  a 
torque  we  have 


dt 


(/«)■ 


(Ill) 


Therefore  torque  equals  the  time  rate  of  change  of  momentum. 
The  following  analysis  proves  that  the  last  statement  is 
true  when  the  moment  of  inertia  varies  with  the  time  as 
well  as  when  it  remains  constant. 

Let  A,  Fig.  122,  represent  a  body,  or  a  system  of  bodies, 
which  is  acted  upon  by  one  or  more  external  torques.  For 
the  sake  of  simplicity  suppose 
the  planes  of  the  torques  to  be 
parallel  to  the  plane  of  the 
paper,  and  the  axis  of  rotation 
to  pass  through  the  point  0  and 
to  be  perpendicular  to  the  plane 
of  the  paper.  Let  dF  be  the 
resultant  force  acting  upon  an 
element  of  mass  dm.  Then  the  moment  of  dF  about  the 
axis  of  rotation  equals  the  product  of  r,  the  distance  of  dm 
from  the  axis,  by  dFpf  the  component  of  dF  perpendicular 
to  r.     Therefore 

dF, 

dmfp 


Fig.  122. 


dG 


dm  • 


1  d 

/■  dt 


(r2*)       [p.  97] 


dm  •-,  (r2«) 
dt 


=  |  (r=,/,„ 


»). 


268  ANALYTICAL  MECHANICS 

Therefore  the  resultant  external  torque  acting  upon  the 
body  is 

or  G=|(/w)'  (III) 

where  /  is  supposed  to  vary  with  the  time.  Equation  (III) 
is  the  genera]  form  of  torque  equation,  of  which  equation 
(V)  of  Chapter  XI  is  a  special  case. 

Introducing  this  expression  of   G  in  the  definition  for 
angular  impulse  we  obtain 


-i 

-I 


Gdt 


=  7o>  -  70co0,  (IV) 

where  h  and  <Do  denote  the  moment  of  inertia  and  the  angular 
acceleration  at  the  instant  t  =  0,  and  I  and  to  those  at  t  —  t. 
Equation  I IV)  is  a  generalization  of  equation  (II).  It  states 
tti.it  angular  impulse  equals  the  change  in  the  angular  momen- 
tum under  nil  circumstances. 

211.  The  Principle  of  the  Conservation  of  Angular  Momen- 
tum.— When  the  resultant  external  torque  acting  upon  a 
body  or  system  of  bodies  vanishes,  it  follows  from  equation 
(III)  that 

|(/»)  =  0, 

and  consequently  7©  =  const.  (V) 

Therefore  //  the  resultant  of  all  the  external  torques  acting 
upon  a  system  vanishes,  the  angular  momentum  of  the  system 


ANGULAR  IMPULSE  AND  ANGULAR  MOMENTUM     269 

remains  constant,  in  direction  as  well  as  in  magnitude.  This 
is  the  principle  of  the  conservation  of  angular  momentum. 

ILLUSTRATIVE  EXAMPLE. 

Discuss  the  effect  of  a  shrinkage  in  the  radius  of  the  earth  upon  the 
length  of  the  day. 

Let  P  and  P'  be  the  lengths  of  the  day  when  the  radius  of  the  earth  is  a 
and  a',  respectively.  Further,  let  a;  and  co'  be  the  corresponding  values 
of  the  angular  velocity  of  the  earth  about  its  axis.    Then 

S-5- 

But  since  the  earth  is  not  supposed  to  be  acted  upon  by  any  external 
torques  its  angular  momentum  remains  constant.     Therefore 

Ico  =  ZV.  (2) 

From  equations  (1)  and  (2)  we  obtain 

El  =  11  =  °11 

PI  a2' 
P-P'  a2  -  a'2 
-P~  =  ^ 

and  *§  =  "±01..**,  (3) 

P  a        a 

where  8P  and  8a  denote  the  diminutions  in  the  length  of  the  day  and  the 
radius,  respectively.  When  8a  is  small  a'  is  very  nearly  equal  to  a,  there- 
fore equation  (3)  may  be  written  in  the  form 

Therefore  the  percentage  diminution  in  the  length  of  the  day  is  twice  as 
large  as  the  percentage  diminution  in  the  radius.  Hence  when  the  radius 
is  diminished  by  1  mile  the  length  of  the  day  is  diminished  by  about  43 
seconds. 

PROBLEMS. 

1.  How  do  the  oceanic  currents  from  the  polar  regions  affect  the  length 
of  the  day? 

2.  A  uniform  rod  of  negligible  diameter  falls  from  a  vertical  position 
with  its  lower  end  on  a  perfectly  smooth  horizontal  plane.  What  is  the 
path  of  its  middle  point? 


270  ANALYTICAL  MECHANICS 

3.  While  passing  through  the  tail  of  a  comet  an  amount  of  dust  of 
mass  hi  settles  uniformly  upon  the  surface  of  the  earth.  Find  the  conse- 
quent change  in  the  Length  of  the  day. 

4.  In  the  preceding  problem  find  the  torque  due  to  the  addition  of 
mass.  Suppose  the  passage  to  take  n  days  and  the  rate  at  which  mass 
is  acquired  to  be  constant. 

6.  A  particle  revolves,  on  a  smooth  horizontal  plane,  about  a  peg,  to 
which  it  i-  attached  by  means  of  a  Btring  of  negligible  mass.  The  string 
winds  around  the  peg  as  the  particle  rotates.  Discuss  the  motion  of  the 
particle. 

6.  A  mouse  is  made  to  run  around  the  edge  of  a  horizontal  circular 
table  which  is  free  to  rotate  about  a  vertical  axis  through  the  center. 
Find  the  velocity  of  the  mouse  relative  to  the  table  which  will  give  the 
latter  20  revolutions  per  minute?  The  table  weighs  2  pounds  and  has  a 
diameter  of  is  inches;  the  mouse  weighs  5  ounces. 

7.  In  the  preceding  problem  find  the  velocity  of  the  mouse  with  re- 
specl  to  the  ground. 

8.  A  cylindrical  vessel  of  radius  a  is  filled  with  a  liquid,  closed  tight, 
and  made  to  rotate  with  a  constant  angular  velocity  oj0 about  its  geomet- 
rical axis,  which  is  vertical.  Suppose  the  frictionaJ  forces  between  the 
inner  surface  of  the  \  esse!  and  the  liquid  and  bet  ween  the  molecules  of 
the  liquid  to  be  small,  yet  enough  to  transmit  the  motion  to  the  liquid  if 
the  rotation  is  kept  up  for  a  long  time.  After  each  particle  of  water  at- 
tains an  angular  velocity  about  the  axis  given  by  the  relation  co  =  co0r  the 
torque  which  kepi  the  angular  velocity  constant  is  stopped  and  the  liquid 
i-  suddenly  fro/en.     What  will  be  1  he  angular  velocity  of  the  system  if 

The  mass  of  the  vessel  is  negligible, 
(b)  The  maSfi  ifl  not   negligible  but   the  thickness  is.     Take  the  ends 
into  account . 

Neither  the  mass  nor  the  thickness  of  the  cylinder  is  negligible. 
I  k)  not  take  the  ends  into  account. 

(d)   In  (c)  take  the  ends  into  account. 

9.  In  the  preceding  problem  suppose  the  distribution  of  the  angular 
velocity  of  the  liquid  about  the  axis  just  before  it  is  frozen  to  be  given 

a—  r 

by  the  relation  w  =  co0e    r    .  where  ;•  is  the  distance  from  the  axis. 


A.\<;i  I.Ai;    IMPULSE  AXD  ANGULAB   MOMENTUM     271 


APPLICATION  TO  SPECIAL  PROBLEMS. 
212.  Ballistic  Pendulum. —A  ballistic  pendulum  is  a  heavy 
target  which  is  used  to  determine  the  velocity  of  projectile-. 
The  target,  which  is  suspended  from  a  horizontal  axis,  is 
given  an  angular  displacement  when  it  receives  the  projec- 
tile. Considering  the  target  and  the  bullet  which  is  projected 
into  it  as  an  isolated  system  we  apply  the  principles  of  the 
conservation  of  energy  and  of  the  conservation  of  angular 
momentum.  Just  before  the  bullet  hits 
the  target  the  angular  momentum  of 
the  system  about  the  axis  is  that  due 
to  the  velocity  of  the  bullet  and  equals 

/'  7,  where  /'  is  the  moment  of  inertia 
b 

of  the  bullet  about  the  axis,  v  is  its 

velocity,  and  b  is  its  distance  from  the 

axis  just  before  it  hits  the  target,  Fig. 

123.     The  bullet  is  supposed  to  hit  the 

target  normally,  when  the  latter  is  in  the 

equilibrium  position,  and  t  o  be  imbed<  led 

in  it.  The  angular  momentum  just  aft  er 

the  bullet  hits  the  target  is  (/  + 1')  u, 

where  /  is  the  moment  of  inertia  of  the  target  and  w  its  initial 

angular  velocity.     Then,  by  the  conservation  of  the  angular 

momentum,  we  have 

(1) 


l'\  =  (/+/')  co.      .-.  V=bl~ 


b        *-    '    -'~  /'      - 

If  we  suppose  the  energy  lost  during  the  impact  to  be  negli- 
gible the  kinetic  energy  of  rotation  just  after  the  bullet  hits 
the  target  equals  the  potential  energy  of  the  system  at  its 
position  of  maximum  angular  displacement.     Therefore 

I  (/  +  /')  co2  =  (M  +  m  >  ga  1 1  -  cos  a),  (2) 

where  .1/  and  m  are  the  masses  of  the  target  and  of  the  bullet , 
respectively,  a  is  the  distance  of  the  center  of  mass  of  the 
system  from  the  axis,  and  a  is  the  maximum  angular  dis- 


272  ANALYTICAL  MECHANICS 

placement.  Eliminating  «  between  equations  (1)  and  (2)  we 
obtain 

v  =  A  V2  ga  (/  +  /')  (M  +  m)  (1  -  cos  a) .  (3) 

The  moment  of  inertia  of  the  target  may  be  determined  by  ob- 
serving the  period  of  oscillation  when  it  is  used  as  a  pendulum. 
It  will  be  shown  later*  that  if  P  denotes  the  period  then 

P-W(i£fcr  (4) 

Eliminating  (/  -f-  /')  between  equations  (3)  and  (4)  we  get 

(5) 


Pabg  (M  +  m)    /!_ 


v 

Pa  by  (M+m)     ■    0a 


id'  2 

But  in  practice  m  is  very  small  compared  with  M,  the  bullet 
is  small  enough  to  be  considered  as  a  small  particle,  and  a  is 
small;  therefore  we  can  neglect  m  in  the  numerator,  substi- 
tute mb'1  for  V,  and  replace  sin  ^  by  ^.     When  these  simpli- 

z        z 

iiiations  are  introduced  into  equation  (5)  we  get 

PagM    2  fas 

V=I*mba--  (6) 

213.  Motion  Relative  to  the  Center  of  Mass. — Suppose  a 
rigid  body  to  have  a  uniplanar  motion.  Let  M  be  the  mass 
of  the  body,  /  its  moment  of  inertia  with  respect  to  an  axis 
perpendicular  to  the  plane  of  the  motion,  Ic  its  moment  of 
inertia  about  a  parallel  axis  through  the  center  of  mass,  and 
a  the  distance  between  the  two  axes.  Then  the  angular 
momentum  about  the  first  axis  is 

7a,  =  (/,.+  Ma2)  co 


=  /,.«  + a- My,1  (M) 

where  I  is  the  velocity  of  the  center  of  mass.     In  the  right 
hand  member  of  the  last  equation  the  first  term  represents 
*  Page  309. 


ANGULAR  IMPULSE  AND  ANGULAR  MOMENTUM     273 


the  angular  momentum  of  the  body  due  to  the  motion  of 
its  particles  relative  to  the  center  of  mass,  while  the  second 
term  represents  the  angular  momentum  of  the  body  due  to 
the  motion  of  its  particles  with  the  center  of  mass.  The 
second  term  depends  upon  the  position  of  the  center  of  mass 
relative  to  the  axis  of  rotation.  The  first  term  does  not  at 
all  depend  upon  this  position.  It  depends  upon  the  distri- 
bution of  the  particles  of  the  body  about  the  center  of  mass. 
The  two  terms  are,  therefore,  independent;  that  is,  if  the 
center  of  mass  of  a  body  is  suddenly  fixed  the  angular  mo- 
ment um  of  the  body  due  to  the  motion  of  its  particles  about 
the  center  of  niass  is  not  at  all  affected.  On  the  other  hand 
if  the  motion  about  the  center  of  mass  is  destroyed  the  angu- 
lar momentum  about  a  given  axis  due  to  the  motion  of  the 
particles  of  the  body  with  the  center  of  mass  is  not  changed. 
In  other  words  motion  about  the  center  of  mass  and  motion  with 
the  center  of  mass  are  distinct  and  independent* 

As  an  illustration  of  this  important 
fact  consider  two  disks,  Fig.  124,  of  equal 
mass,  radius,  and  thickness,  which  have 
equal  and  opposite  angular  velocities 
about  a  common  axle,  and  which  move 
with  the  axle  in  a  direction  perpendicular 
to  it.  Suppose  each  of  the  disks  to  have 
two  similarly  placed  holes,  as  shown  in 
the  figure,  so  that  they  can  be  made  one 
solid  piece  by  dropping  a  pin  in  each  pair 
of  holes  when  they  are  in  line.  If  the 
rotational  motion  is  stopped  by  dropping 
the  pins  into  the  holes,  the  motion  of  the 
axle  goes  on  as  if  aotbing  had  happened. 
On  the  other  hand  if  the  motion  with  the 
axle  is  changed  or  even  stopped,  the  rotations  of  the  disks  about  the  axle 
are  not  at  all  disturbed. 


Fig.  124. 


This  result  holds  true  for  all  bodies  and  systems,  whether  rigid  or  not. 


274 


ANALYTICAL  MECHANICS 


ILLUSTRATIVE    EXAMPLE. 

A  uniform  circular  hoop  rotates  about  a  peg  on  a  perfectly  smooth 
horizontal  plane;  find  the  angular  velocity  of  .the  hoop  if  the  peg  is  sud- 
denly removed  and  simultaneously  another  peg  is  introduced,  about 
which  it  begins  to  rotate. 

Let  0  and  0',  Fig.  125,  be  the  positions  of  the  first  and. second  peg, 
respectively.     The  circle  in  continuous  line  may  be  considered  to  repre- 
sent the  position  of  the  hoop  just 
before    it    stops   rotating   about  0 
and  just  after  it  begins  to  rotate 
about  0'. 

The  only  force  which  comes  into 
play  when  the  hoop  strikes  the  peg 
<>'  passes  through  0',  hence  it  pro- 
duces no  effect  upon  the  angular 
momentum  about  0'.  Therefore 
the  angular  momentum  about  0' 
ju-t  after  the  hoop  strikes  the  peg 
equals  the  angular  momentum  just 
before.  The  angular  momentum 
after  the  hoop  begins  to  rotate  about  0'  is 

H'0'  =  Ico'  =  2  ma2u', 

where  //,'.  is  the  angular  momentum  and  u'  the  angular  velocity  about 
the  <>',  in  the  mass,  and  a  the  radius  of  the  hoop. 

The  angular  momentum  about  0'  just  before  the  hoop  begins  to  rotate 
about  ()'  equals  the  angular  momentum  of  the  hoop  due  to  the  motion  of 
the  hoop  about  its  <,r<'omet rical  axis  plus  its  angular  momentum  due  to 
its  motion  with  its  center  of  mass.     Therefore 
IK'  =  ^|  mo  •  a  cos  a 
=  ma*a  +  ma2o)  cos  a 
=  ma2u)  (1  -f  cos  a), 

where  u  ifl  the  angular  velocity  about  the  peg  0,  and  a  the  angle  which 
the  arc  ()()■  subtends  at   the  center  of  the  hoop.      Bui  since 

//V  =  Ho', 
2  rnaaa>',=  mafa  (1  -f  cos  a) 
and  „'  =  L±_£os«^ 

,,,1  +  coso^ 


ANGULAR  IMPULSE  AND  ANGULAR  MOMENTUM     275 

where  v  is  the  linear  velocity  of  the  center  of  the  hoop  while  the  latter 
rotates  about  o,  and  v'  the  velocity  afterwards. 

Discussion.  —  When  a  =  0,  that  is,  when  the  two  pegs  coincide, 

co'  =  co  and  v'  =  v,  as  they  should.     When  a  —  -z,  o>'=  -,  v'=  -•      When 

a  =  7T,  a/  =  0  and  v'  =  0,  that  is,  the  hoop  comes  to  rest. 


PROBLEMS. 

1.  A  rod  of  negligible  transverse  dimensions  and  length  I  is  moving  on 
a  smooth  horizontal  plane  in  a  direction  perpendicular  to  its  length.  Show 
that  if  it  strikes  an  obstacle  at  a  distance  a  from  its  center  it  will  have 

an  angular  velocity  equal  to   -=j-,  where  v  is  its  linear  velocity  before 

meeting  the  obstacle. 

2.  A  uniform  circular  plate  is  turning  about  its  geometrical  axis  on  a 
smooth  horizontal  plane.  Suddenly  one  of  the  elements  of  its  lateral  sur- 
face is  fixed.     Show  that  the  angular  velocity  after  fixing  the  element 

equals  ^ ,  where  a;  is  the  angular  velocity  before  fixing  it. 
o 

3.  A  circular  plate  which  is  rotating  about  an  element  of  its  lateral 
surface  is  made  to  rotate  about  another  element  by  suddenly  fixing  the 

second  and  freeing  the  first.     Show  that  a/  = co,  when1  w  and 

o 

w'  arc  the  values  of  the  angular  velocity  of  the  plate  before  and  after  fixing 

the  second  element,  and  a  is  the  angular  separation  of  the  two  elements 

when  measured  at  the  center  of  the  disk. 

4.  Three  particles  of  equal  mass  are  attached  to  the  vertices  of  an 
equilateral  triangular  frame  of  negligible  mass.  Show  that  if  one  of  the 
vertices  is  fixed  while  the  frame  is  rotating  about  an  axis  through  the 
center  of  the  triangle  perpendicular  to  its  plane  the  angular  velocity  is 
not  changed. 

5.  A  square  plate  is  moving  on  a  smooth  horizontal  plane  with  a  veloc- 
ity vat  right  angles  to  two  of  its  sides.  Find  the  velocity  with  which  it 
will  rotate  if 

(a)  one  of  its  corners  is  suddenly  fixed; 

(b)  the  middle  point  of  one  of  its  sides  is  fixed. 

6.  A  uniform  rod  of  negligible  transverse  dimensions  is  rotating  about 
a  transverse  axis  through  one  end.  Find  the  angular  uelocity  with  which 
it  will  rotate  if  the  axis  is  suddenly  removed  and  simultaneously  a  parallel 

axis  is  introduced  through  the  center  of  mass  of  the  rod. 

7.  An  equilateral  triangular  plate  i-  rotating  about  an  axis  through 


276 


ANALYTICAL  MECHANICS 


one  of  the  vertices  perpendicular  to  the  plane  of  the  plate.  Find  the 
resulting  angular  velocity  due  to  a  sudden  removal  of  the  axis  and  a  simul- 
taneous introduction  of  a  parallel  axis  through  the  center  of  mass. 

8.    In  the  preceding  problem,  suppose  the  new  axis  to  pass  through  one 
of  the  other  two  vertices. 


Fig.  126. 


214.  Reaction  of  the  Axis  of  Rotation.— Suppose  B,  Fig.  126, 
to  be  a  rigid  body  free  to  rotate  about  a  fixed  axis  through 
the  point  0,  perpendicular  to  the 
plane  of  the  figure.  If  an  ex- 
ternal force  F  is  applied  to  the 
body  a  part  of  its  action  is,  in  gen- 
et al,  transmitted  to  the  axis  of 
rotation.  This  results  in  the  re- 
action, R,  of  the  axis,  which  we 
will  investigate.  For  the  sake  of 
simplicity  suppose  F  to  lie  in  the 
plane  which  passes  through  the 
center  of  mass,  c,  perpendicular  to  the  axis. 

Since  F  and  R  are  supposed  to  be  the  only  external  forces 
acting  upon  the  body,  then  by  equation  (VIII)  of  p.  242 

rav  =  F  +  R,  (1) 

where  tf  is  the  acceleration  of  the  center  of  mass.  If  F„  and 
F  denote  the  components  of  F  along  and  at  right  angles  to 
the  line  Oc,  respectively,  and  P  and  Q  the  components  of  R 
along  the  same  directions,  equation  (1)  may  be  resolved 
into  the  following  component-equations: 

mfn  =  Fn  +  P, 

ml  =  FT  +  Q, 


where  ?„  and  ?r  are  the  components  of  v. 
of  the  center  of  mass  is  a  circle 


(2) 
(3) 
But  since  the  path 


and 


Jn      a 

fr=r 


Qco, 


ANGULAR  IMPULSE  AND  ANGULAR  MOMENTUM     277 

where  a  is  the  distance  of  the  center  of  mass  from  the  axis 
and  co  the  angular  velocity  of  the  body.  Making  these  sub- 
stitutions in  equations  (2)  and  (3)  and  solving  for  P  and  Q 
we  obtain 

P=-Fn  +  mau2,  (VII) 

Q  =  -  Ft  +  maco.  (VIII) 

The  magnitude  and  the  direction  of  R  are  given  by  the 
relations 


r  =  Vp*  +  Q 


and  tan</>=^> 

where  <t>  is  the  angle  R  makes  with  the  line  Oc. 

ILLUSTRATIVE  EXAMPLE. 

A  uniform  rod,  which  is  free  to  rotate  about  a  horizontal  axis  through 
one  end,  falls  from  a  horizontal  position.  Find  the  reaction  of  the  axis  at 
any  instant  of  its  fall. 

Evidently  Fn  =  —  mg  cos  0. 

FT  =  —  mg  sin  0. 

The  negative  sign  in  the  first  equation  is  due  to  the  fact  that  in  equa- 
tion (VII)  Fn  is  supposed  to  be  directed  towards  the  axis,  while  mg  cos0 
is  directed  away  from  the  axis.  The  negative  sign  in  the  second  equation 
is  due  to  the  fact  that  6  is  measured  in  the  counter-clockwise  direction, 
while  mg  sin  6  points  in  the  opposite  direction. 

Substituting  these  values  of  Fn  and  l<\  in  equations  (VII)  and  (VHI), 
we  obtain 

P  =  mg  cos  6  +  mr/or, 

Q  —  mg  sin  6  -f  mnic. 

But  by  the  conservation  of  energy 

\  la*  =  mga  cos  6, 

where  a  is  one-half  the  length  of  the  rod.     Therefore 

,      2  mqa        fl      3o        a 
co2  =  — Is—  cos  0  =  —*•  cos  0 
/  2 " 

and  o>  =  —  — ^  sin  0. 

4a 


278  ANALYTICAL   MECHANICS 

Making  these  substitutions 

P  =  5  mg  cos  6. 

Q  =  *  >»g  sin  6. 

R  =  1J1?-  Vl+99cos20. 
4 

tan  0  =  iV  tan  8. 

Discussion.  — The  reaction  and  its  direction  are  independent  of  the 
length  of  the  rod.  When  6  =  0,  Q  =  0  and  R  =  P  =  f  mp.  In  other 
words  at  the  instant  when  the  rod  passes  the  lowest  point  the  force  on  the 

axis  ia  :  times  as  large  as  the  force  when  rod  hangs  at  rest.     When  8  =  -> 

P  =  0  and  R  =Q  =  \  mg.  If  the  rod  is  held  in  a  horizontal  position  by 
supporting  the  free  end  the  reaction  of  the  axis  is  §  mg.  But  as  soon  as 
the  support  is  removed  from  the  free  end  the  reaction  on  the  axis  is  changed 
from  \  mg  to  \  mg. 

PROBLEMS. 

1.  A  uniform  rod'  which  is  free  to  rotate  about  a  horizontal  axis  falls 
from  the  position  of  unstable  equilibrium.     Find  the  reaction  of  the  axis. 

2.  In  the  preceding  problem  find  the  position  where  the  horizontal 
component  of  the  reaction  is  a  maximum. 

3.  A  uniform  rod  which  is  free  to  rotate  about  a  horizontal  axis  falls 
from  a  horizontal  position.  Show  that  the  horizontal  component  of  the 
reaction  is  greatest  when  the  rod  makes  45°  with  the  vertical. 

4.  A  cube  rotates  about  a  horizontal  axis  which  coincides  with  one  of 
its  edges.     If  at  the  highest  position  it  barely  completes  the  revolution, 

show  that  P  =  :!  ~  ,r;>(""s  6  W  and  Q  =  ^~  W,  where  W  is  the  weight  of 

the  cube. 

5.  A  cube  which  is  free  to  rotate  about  a  horizontal  axis  through  one 
of  its  edges  Starts  to  fall  when  its  center  is  at  the  same  level  as  the  axis  of 
rotation.      Kind  the  reaction  of  the  axis. 

6.  Show  that  if  the  body  of  5 2]  1  is  a  particle  connected  to  the  axis 
with  a  massless  rod  the  reaction  perpendicular  to  the  rod  vanishes. 

7.  Consider  the  reactions  of  the  axis  when  the  latter  passes  through 

the  center  of  mass  of  the  rigid  body. 

8.  A  circular  plate  is  free  to  rotate  about  a  horizontal  axis  which  forms 

one  of  the  elements  of  its  cylindrical  Burface.    The  plate  is  let  fall  from 
:tion  when  its  center  of  mass  is  vertically  above  the  axis.    De- 
termine the  reaction  of  the  axis  at  0  =  *  and  at  6  =  0. 


ANGULAR  IMPULSE  AND  ANGULAR   MOMENTUM     _'7!> 


9.  A  hoop  barely  completes  rotations  about  a  horizontal  axis  which 
passes  through  its  rim  and  is  perpendicular  to  its  plane      Determine 

the  reaction  of  the  axis  at  the  lowest  and  the  bighesl  positions. 

10.  A  uniform  rod  which  rotates  about  a  horizontal  axis  through  one 

end  has  four  times  as  much  kinetic  energy  as  it  has  potential  energy  at  the 
instant  it  passes  the  highest  point.  Find  the  reaction  of  the  axis  when 
the  rod  is 

(a)  at  the  highest  position; 

(b)  horizontal; 

(c)  at  the  lowest  position. 

215.  Impulsive  Reaction  of  an  Axis.  Center  of  Percussion. — 
If  a  rigid  body  which  is  free  to  rotate  about  a  fixed  axis  is 
so  struck  that  no  impulse  is  imparted  to  the  axis  during  the 
blow,  any  point  of  the  line  of  action  of  the  blow  is  called  a 
center  of  percussion  for  that  axis.  It  is  evident  that  if  the 
axis  be  removed  and  the  blow  applied  at  a  center  of  percus- 
sion which  corresponds  to  the  removed  axis,  the  body  will 
rotate  as  if  the  axis  were  not  removed.  The  axis  about 
which  a  free  rigid  body  rotates  when 
it  is  given  a  blow  is  called  the  axis 
of  spontaneous  rotation. 

Suppose  the  rigid  body  of  Fig. 
127  to  be  free  to  rotate  about  an 
axis  through  0  perpendicular  to  the 
plane  of  the  figure.  For  the  sake 
of  simplicity  suppose  the  blow  to  be 
applied  in  such  a  direction  that  it 
tends  to  produce  rotation  only  about 
the  given  axis.  Let  L  denote  the 
linear  impulse  of  the  blow  and  L' 

the  impulse  given  to  the  body  by  the  reaction  of  the  axis 
of  rotation.  Then  by  the  conservat  ion  of  linear  momentum 
the  linear  momentum  of  the  body  must  be  equal  to  the 
impulse  given  to  it  by  the  blow  and  by  the  reaction  of  the 
axis.     Therefore 

mv  =  L+l',  (1) 


280  ANALYTICAL  MECHANICS 

where  m  is  the  mass  of  the  body  and  v  the  velocity  of  its 
center  of  mass.  But  by  the  conservation  of  angular  momen- 
tum the  angular  momentum  of  the  body  about  the  axis  after 
the  blow  must  equal  that  of  the  blow  itself.     Therefore 

la  =  Lb,  (2) 

where  /  is  the  moment  of  inertia  of  the  body,  «  its  angular 
velocity  and  b  the  distance  of  the  line  of  action  of  the  blow 
from  the  axis. 

Eliminating  L  between  equations  (1)  and  (2)  and  solving 
for  L'  we  obtain 

L  =  mv — —  >  (3) 

b 

*-£)»,  ax) 

where  a  is  the  distance  of  the  center  of  mass  from  the  axis  of 
iot  at  ion.  Equation  (IX)  gives  the  impulse  produced  by  the 
reaction  of  the  axis. 

If  tho  blow  is  applied  at  a  center  of  percussion  V  =  0. 
Therefore 

ma  —  7=0 
0 

and  b= (X) 

ma 

PROBLEMS. 

1.  A  square  plate  is  moving  on  a  smooth  horizontal  plane  with  two  of 

parallel  to  the  direction  of  motion.     Find  the  angular  velocity 
with  which  it  will  rotate,  also  the  impulsive  reaction  of  the  axis, 

(a)  if  one  of  the  corners  is  fixed; 

(b)  if  the  middle  poinl  of  one  of  the  sides  is  fixed. 

2.  Aii  equilateral  triangular  plate  is  moving  on  a  smooth  horizontal 
plane  in  a  direction  perpendicular  to  one  <>f  its  sides.  Find  the  resulting 
angular  velocity,  also  the  impulse  given  by  the  axis, 

if  one  of  its  corners  is  fixed  ; 

if  the  middle  point  of  one  of  its  sides  is  fixed. 

3.  A  hoop  is  moving  on  a  smooth  horizontal  plane  with  its  axis  perpen- 


ANGULAR  IMPULSE  AND  ANGULAR  MOMENTUM     281 

dicular  to  the  plane.  Suppose  a  point  on  it  to  be  fixed  and  find  expres- 
sions for  the  resulting  angular  velocity  and  impulse  imparted.  Discuss 
the  expressions  for  special  positions  of  the  fixed  point. 

4.  While  a  circular  plate  is  moving  on  a  smooth  horizontal  plane  one 
of  the  elements  of  its  lateral  surface  is  fixed.  Find  expressions  lor  the 
resulting  angular  velocity  and  the  impulse  given  by  the  axis.  Discuss 
the  results  for  special  positions  of  the  axis  of  rotation. 

6.  A  uniform  rod  lies  on  a  smooth  horizontal  plane.  Where  must  a 
blow  he  struck  so  that  it  rotates  about  one  end? 

6.  In  the  preceding  problem  can  the  rod  be  made  to  rotate  about  its 
middle  point  by  a  single  blow? 

7.  A  circular  plate  which  lies  on  a  smooth  horizontal  plane  is  struck  so 
that  it  rotates  about  one  of  the  elements  of  its  lateral  surface  as  an  axis. 
Find  the  position  where  the  blow  is  applied. 

8.  Find  the  center  of  percussion  of  a  hoop  which  is  free  to  rotate 
about  an  axis  perpendicular  to  its  plane. 

9.  How  must  a  triangular  plate,  placed  on  a  smooth  horizontal  plane, 
be  struck  so  that  it  may  rotate  about  one  of  its  vertices? 

( ;EXERAL  PROBLEMS. 
1.  Two  particles  of  equal  mass  are  connected  by  a  string  of  length  I 
and  of  negligible  mass  and  placed  on  a  smooth  horizontal  table  so 
that  one  of  the  particles  is  near  an  edge  of  the  table  and  the  string  is 
stretched  at  right  angles  to  the  edge.  The  particle  near  the  edge  is  given 
a  small  displacement  so  that  it  begins  to  fall.  Show  that  the  interval  of 
time  between  the  instant  at  which  the  second  particle  leaves  the  table 
and  the  instant  at  which  the  string  occupies  a  horizontal  position  is 
given  by 


\l\ 


2.  A  uniform  bar  of  negligible  cross-section,  which  is  rotating  on  n 
smooth  horizontal  plane  about  a  vertical  axis,  strikes  an  obstacle  and 
begins  to  rotate  in  the  opposite  direction.  If  L  and  L'  denote  the  impulses 
given  by  the  collision  to  the  axis  and  the  obstacle,  respectively,  u  and  a/ 
the  angular  velocities  of  the  bar  before  and  after  the  collision,  /  the  length 
and  ///  the  mass  of  the  bar,  and  a  the  distance  of  the  obstacle  from  the  axis, 
show  that 

(a)  a/  =  cu>; 

n\   t  /i  j    ,1(7!  -  6a) 

(b)  L  =  m  (1  +  e)  — — -«; 

12a 

(c)  L'  =  m(i  +  e)^«. 


282  ANALYTICAL  MECHANICS 

3.  A  circular  table  is  perfectly  free  to  rotate  about  a  vertical  axis 
through  its  center.     Show  t  hat  if  a  man  walks  completely  around  the  edge 

of  the  table  the  latter  turns  through  an  angle  of  ^  •  2  tt,  where  m 

and  M  an-  the  masses  of  the  table  and  of  the  man,  respectively. 

4.  A  circular  plate  is  rotating  about  its  axis,  which  is  vertical,  with  an 
angular  velocity  w  and  is  moving  on  a  smooth  horizontal  plane  with  a  linear 
velocit:  Find  the  angular  velocity  it  will  have  if  one  of  the  elements 
of  il-  lateral  surface  is  suddenly  fixed,  and  determine  the  impulse  given 
;  axis  of  rotation.  Discuss  the  results  for  special  positions  of  the 
fixed  axis. 

6.  A  uniform  rod  strikes  at  one  end  againsl  an  obstacle  while  falling 
transversely.  Show  that  the  impulse  which  the  obstacle  receives  will 
be  one-half  that  which  it  would  have  received  if  the  other  end  of  the 
rod  had  struck  an  obstacle  simultaneously  with  the  first. 

6.  A  particle  is  projected  into  a  tube  which  is  bent  to  form  a  circle  and 
is  lying  on  a  smooth  horizontal  table.  If  the  inner  surface  of  the  tube  is 
perfectly  smooth,  show  that  the  center  of  mass  of  the  two  moves  in  the 
direction  of  projection  of  the  particle  with  a  velocity  of; —  ..  v,  while 
the  particle  and  the  center  of  the  tube  describe  circles  about  it  with  an 

angular  velocity  -.  where  .1/  is  the  mass  and  a  the  radius  of  the  tube, 
a 

while  m  is  the  mass  and  v  the  velocity  of  projection  of  the  particle. 

7.  bind  the  direction  and  point  of  application  which  an  impulse  must 
have  in  order  to  make  a  sphere  rotate  about  a  tangent. 

8.  A  uniform  rod  which  is  rotating  on  a  smooth  horizontal  plane  about 
a  pivot  through  its  middle  point  breaks  into  two  equal  parts.     Determine 

the  subsequent    motion. 

9.  A  uniform  rod  rotates  on  a  smooth  horizontal  plane  about  a  pivot. 
What  will  be  the  motion  when  the  pivot  breaks? 

10.  A  uniform  rod  falls  from  a  position  where  its  lower  end  is  in  con- 
tact with  a  rough  horizontal  plane  with  which  it  makes  an  angle  a.     Show 

that  when  it  becomes  horizontal  its  angular  velocity  is  y— ^— ,  where 

/  i-  the  length  of  the  rod. 

11.  Show  that  in  problem  (10)  the  angular  velocity  will  be  the  same 
when  the  horizontal  plane  i-  smooth. 

12.  A  uniform  rod  which  lies  on  a  smooth  horizontal  plane  is  struck 

id.  transversely.    Show  that  the  energy  imparted  equals  f  of  the 
which  would  have  been  given  to  the  bar  by  the  same  blow  if  the 
end  of  the  bar  were  fixed. 


CHAPTER   XIV. 

MOTION   OF   A   PARTICLE   IN   A   CENTRAL   FIELD 
OF   FORCE. 

216.  Central  Field  of  Force.  —  A  region  is  called  a  central 
field  of  force  when  the  intensity  of  the  field  at  every  point 
of  the  region  is  directed  toward  a  fixed  point.  The  fixed 
point  is  called  the  center  of  the  field.  The  force  which  a 
particle  experiences  when  placed  in  a  central  field  of  force 
is  called  a  central  force. 

217.  Equations  of  Motions.  —  Consider  the  motion  of  a  par- 
ticle which  is  projected  into  a  central  field  of  force.  It  is 
evident  from  symmetry  that  the  path  will  lie  in  the  plane 
determined  by  the  center  of  the  field  and  the  direction  of 
projection.  The  expressions  for  the  radial  and  transverse 
components  of  the  acceleration  are,  according  to  the  results 
of  §90, 

f  _d2r        /dey 
Jr~dc>  ~r[dt)' 

When  the  center  of  the  field  is  chosen  as  the  origin  the  force 
acts  along  the  radius  vector.  Therefore  the  transverse  ac- 
celeration vanishes.-  Suppose  the  force  and  the  acceleral  ion 
to  be  functions  of  the  distance  of  the  particle  from  the  cen- 
ter, then  the  last  two  equations  become 

dV  _ 

dt2 

d 

dt 

283 


(r*«)  =  0.  (II) 


284  ANALYTICAL  MECHANICS 

where  -/  (r)  is  the  total  acceleration.  The  negative  sign  in 
the  right-hand  member  of  equation  (I)  indicates  the  fact 
that  the  acceleration  is  directed  toward  the  center,  while 
the  radius  vector  is  measured  in  the  opposite  direction. 
Equations  (I)  and  (II)  are  the  differential  equations  of  the 
motion  of  a  particle  in  a  central  field  of  force. 

218.  General  Properties  of  Motion  in  a  Central  Field. — 
Integrating  equation  (II)  we  get 

r2u=/*,  (III) 

where  h  is  a  constant.  The  following  properties,  which  are 
direct  consequences  of  equation  (III),  are  common  to  all 
motions  in  central  fields  of  force. 

(1)  The  radius  vector  sweeps  over  equal  areas  in  equal 
intervals  of  time. 

When  the  radius  vector  turns  through  an  angle  dd  it  sweeps 
over  an  area  equal  to  \r>rdd;  therefore  the  rate  at  which 
the  area  is  described  equals 

1  J      1.        1 , 

-r-  —  =  -r-co  =  -  h  =  constant. 

2  dt      2  2 

(2)  The  angular  velocity  of  the  particle  varies  inversely 
as  the  Bquare  of  the  distance  of  the  particle  from  the  center 
of  force.     This  is  evident  from  equation  (III). 

(3)  The  linear  velocity  of  the  particle  varies  inversely  as 
the  Length  of  the  perpendicular  which  is  dropped  upon  the 
direction  of  the  velocity  from  the  center  of  force. 

It  was  shown  on  page  87  that 

VCOSd) 


where  v  is  the  linear  velocity  and  </>  the  angle  which  the 

velocity  makes  with  a  line  perpendicular  to  the  radius  vector. 

i  denote  the  length  of  the  perpendicular  dropped  from 


MOTION  OF  A  PARTICLE  285 

the  center  of  force  upon  the  direction  of  the  velocity;  fcheD 
it  is  evident  from  Fig.  128  that 

cos<j>=  '— 
r 

Substituting  this  value  of  cos  <j> 
in  the  preceding  equation  we 
obtain 

pv 

(x)  =    —  J 

r-  Fig.  128. 

or  v=  —  =  -•  (1\) 

V       P 

(4)  The  angular  momentum  of  the  particle  with  respect 
to  the  center  remains  constant. 

This  result  is  obtained  at  once  by  multiplying  both  sides 
of  equation  (III)  by  m,  the  mass  of  the  particle.     Thus 

mr2w  =  mh, 

but  mr2co  =  Iu. 

Therefore  loo  =  mh  =  constant. 

219.  Equation  of  the  Orbit.  —  The  general  equation  of  the 
orbit  is  found  by  eliminating  t  between  equations  (I)  and 
(III).  The  analytical  reasoning  which  follows  does  not  need 
further  explanation: 

^L  _  dr  d$  _     dr 

dt~  dd'  dt~  U  dd    . 


=  h  dr 
r2  dd 

-  3 

dd 

hdu 

=  -  h  — f 
dd 


[by  (Hi)] 


286 


ANALYTICAL  MECHANICS 


where  u  =  -. 

Therefore 

d-r 
dt2 

,  d2u  dd 
~    do1 '  dt 

_ 

,„  0d2u 

—  h-U'—r-- 

Substituting  this  value  of  -^and  the  value  of  —  ,  which  may 
be  obtained  from  equation  (III),  in  equation  (I),  we  have 


dd*-  h*u* 


(V) 


for  the  equation  of  the  orbit.     When  the  law  of  force  is  given 
known  and  the  orbit  is  determined  by  equation  (V). 
On  the  other  hand  if  the  orbit  is  given  equation  (V)  deter- 
mines the  law  of  force.     Thus,  if  F  denotes  the  force, 
F=-mj  (r) 


=  —  mh 


00/      ,  d2u\ 


(VI) 


II.USTI! atiyi:    EXAMPLE. 

A  particle  describee  a  circle  in  a 
cent  nil  field  of  force.  Determine  the 
law  of  force  if  the  center  of  the  field 
lie-  on  the  path. 

Taking  the  center  as  the  origin, 
29,  and  the  diameter  through 
the  origin  as  the  axis  and  referring 
the  circle  to  polar  COdrdinates  we  ob- 
tain 

r  =  2  a  cos  6, 
»  -  o— l—. 

2  a  cos  0 
fur  the  equation  of  the  orbit.     Differentiating  the  last  equation 

.(_! L\ 

dB%     \acos»0     2  a  cos 0/ 

8  0«M«  -  U. 


(2) 


MOTION  OF  A  PARTICLE  287 

Substituting  in  equation  (VI)  from  equations  (1)  and  (2)  we  get 

F  =  -«^a.  (3) 

Therefore  the  force  varies  inversely  as  the  fifth  power  of  the  distance  from 
the  center  i »f  force.  The  aegative  sign  in  the  second  member  of  equation 
(3)  shows  that  the  force  is  directed  towards  the  origin;  in  other  words, 
it  is  an  attractive  force. 

PROBLEMS. 

1.  Show  that  if  a  particle  describes  the  reciprocal  spiral  rd  =  a  in  a 
central  field  of  force,  the  force  is  attractive  and  varies  inversely  as  the  cube 
of  the  distance  from  the  origin,  which  is  the  center  of  attraction. 

2.  Show  that  if  a  particle  describes  the  logarithmic  spiral  r  =  eae  in 


a  central  field  of  force,  the  expression  for  the  force  is  F 


ih2  (a  +  1) 


r8 

3.  A  particle  moves  in  a  central  field  of  force  where  the  force  is  away 
from  the  center  and  is  proportional  to  the  distance.  Show  that  the  orbit 
is  a  hyperbola. 

4.  Show  that  in  the  preceding  problem  the  radius  vector  sweeps  over 
equal  areas  in  equal  intervals  of  time. 

5.  A  particle  describes  an  ellipse  in  a  field  of  force  the  center  of  which 
is  at  the  center  of  the  ellipse.  Show  that  the  force  varies  directly  as  the 
distance  and  is  directed  towards  the  center. 

6.  In  the  preceding  problem  show  that  the  radius  vector  sweeps  over 
equal  areas  in  equal  intervals  of  time. 

7.  A  particle  describes  an  ellipse  in  a  field  of  force,  the  center  of  which 
is  at  one  focus.  Show  that  the  force  is  towards  the  center  of  force,  and 
is  inversely  proportional  to  the  square  of  the  distance. 

220.  Motion  of  Two  Gravitating  Particles.  —  Suppose  two 
particles  of  masses  m  and  .1/  to  move  under  the  action  of 
their  mutual  gravitational  attraction,  as  in  the  case  of  the 
sun  and  the  earth  or  the  earth  and  the  moon.  Then  if  r  is 
the  distance  between  the  centers  and  y  the  gravitational  con- 
stant the  mutual  force  of  attraction  is 

•  mM 

F=-y-   • 

In  order  to  fix  our  ideas  let  .1/  !><■  the  mass  of  the  sun  and 
m  the  mass  of  the  earth.     Then  the  bud   gives  the  earth 


288  ANALYTICAL  MECHANICS 

an  acceleration  -  y  -j,  while  the  earth  imparts  to  the  sun 

an  acceleration  equal  to  7  —  • 

r~ 

Suppose  at  any  instanl   we  impart  to  both  the  sun  and 
the  earth  a  velocity  equal  and  opposite  to  that  of  the  sun 

and  apply  an  acceleration -.     This  will  bring  the  sun  to 

rest  and  keep  it  at  rest,  without  altering  the  motion  of  the 
earth  relative  to  the  sun.*      This  reduces  the  problem  of 
the  motion  of  the  earth  to  that  of  a  particle  moving  in  a 
central  field  of  force  where  the  acceleration  is 
.,  s  M         m 

7(0  =  -7-7  -y-z 


(VII) 
(VIII) 


Substituting  from  equation  (VIII)   in  equation   (V)   we 
obtain 

d-u   .  fi  ,iS 

^  +  "  =  p  (I) 

for  the  cqiiation  of  the  orbit.     Let  u' =  u—  ~,  then  the 
equation  of  the  orbit  takes  the  form 

du' 

In  onh-r  to  integrate  equation  (2)  let  v  =  — , 

ud 

•  The  acceleration  of  a  particle  relative  to  another  moving  particle  is 
found  t>y  adding  the  negative  of  the  acceleration  of  the  second  particle  to 


r-          r- 

M+m 
-      T      r»     - 

or 

/«-£ 

and 

IF— M^. 

r- 

where 

m=  7  (M  +  m). 

MOTION  OF  A  PARTICLE 

then 

d2u'  _  dv  _  dv 
dd2  ~  dd  ~  du'  ' 

and 

y-rr  +  tt'  =o. 
du 

2SD 


Separating  the  variable  and  integrating 
v-  =  A2  —  u'2. 


"  dd 

'A2  -  u'2. 

Integrating  again 

-i  u 
cos  1-- 

=  6  +  8 

or             u'  =  A 

cos  (6+8). 

Let           u'  =  A 

whenfl  =0, 

then     8  -- 

=  0. 

Therefore 

u'  = 

A  cos  6 

(3) 

is  a  solution  of 
in  equation  (3), 

equation  (2).     Substituting  the  value 

of  u' 

-fi 

+  A  cos  6 

(4) 

and  replacing  u 

by  its  value 

ep 

(5) 

T  — 

1 

-  e  cos  6 

where 

e=  — 

ir.\ 
—,P=- 

1 
a' 

(6) 

Equation  (5)  is  the  well-known  equation  of  a  conic  section. 
Therefore  the  orbit  is  a  conic  section  with  an  eccentricity 

equal  to • 

The  expression  for  the  velocity  at  any  point  of  the  orbit 

may  be  obtained  by  Bubstil  uting  the  values  of  rand  ',  .  which 

at        dt 


290  ANALYTICAL  MECHANICS 

may  be  obtained  from  equations  (5)  and  (III),  respectively, 
in  equation  (3)  of  page  82.     Thus 


.,     /r//V  ,    o/ddY 


=  —  sin20+—  • 
p-               r- 

(7) 

Eliminating 

sin2 

d  between  equations  (5)  and  (7) 

^2=Jr2(e2-1)  +  — 2 

e2p2                                epr 

"«V             )+   r  ' 

we 

get 

(8) 

where 

k=  —  =  const. 

ep 

Therefore      v2-  —  =  ^{e2-l)  =  const.  (9) 

r       L-p- 

221.  Conditions  which  Determine  the  Type  of  the  Orbit.  — 
Suppese  a  gravitating  body  to  be  projected  into  the  field  of 
another  gravitating  body,  which  acts  as  the  center  of  force; 
then  the  type  of  the  orbit  is  determined  by  the  initial  con- 
dition-, thai  is,  the  magnitude  and  the  direction  of  the 
velocity  of  projection  and  the  distance  of  the  particle  from 
the  center  of  force  at  the  instant  of  projection.  Substituting 
the  initial  values  of  v  and  r  in  equation  (9)  and  rearranging 
we  obtain  the  following  expression  for  the  eccentricity: 

The  character  of  the  orbil  is  determined  by  the  value  of 
the  factor  in  the  parent  heses  of  equation  (10).  When  it  van- 
ishes i  is  one,  therefore  the  orbit  is  a  parabola;  when  it  is 
[fi  less  than  one,  therefore  the  orbit  is  an  ellipse; 
and  when  it  is  positive  e  is  greater  than  one,  therefore  the 
orbit  i-  a  hyperbola.  We  have,  therefore,  the  following 
criteria: 


:(*-$  ao) 


Case      I.     The  orbit  is  a  parabola,  when  vf 
Case    II.     The  orbit  is  an  ellipse,     when  r02< 


MOTION  OF  A  PARTICLE  291 

2k 

rQ  ' 
2k 

rQ  ' 

2  k 
Case  III.     The  orbit  is  a  hyperbola,  when  v02> 

7*0 

The  general  expression  for  the  velocity,  which  is  given  by 
equation  (9),  may  be  put  in  the  following  special  forms: 

2  A* 
I.     v2  =  — ,  when  the  orbit  is  a  parabola. 

II.     v~  —  k( J,  when  the  orbit  is  an  ellipse. 

III.     v2  =  kl-  +  -J,  when  the  orbit  is  a  hyperbola. 

The  quantity  a  is  the  length  of  the  semi-transverse  axis. 

222.  Velocity  from  Infinity.  —  The  velocity  which  the  par- 
ticle acquires  in  falling  towards  the  center  from  a  point 
infinitely  distant  from  the  center  is  called  the  velocity  from 
infinity.  This  velocity  may  be  computed  from  the  energy 
equation.     Thus 

PrFdr 


\  mv2  =  /  1 

-I 


r 

Therefore  v2=—- 

r 

But  the  last  equation  is  identical  with  the  relation  which 
gives  the  velocity  of  a  panicle  moving  in  a  parabolic  path, 
therefore  if  a  particle  describes  a  parabolic  orbil  its  velocity 
atany  point  of  it>  <>rl>it  is  equal  to  the  velocity  it  would  have 
acquired  if  it  had  started  from  infinity  and  arrived  at  that 


292  ANALYTICAL  MECHANICS 

point  of  the  field  of  force.  This  fact  enables  us  to  state  the 
conditions  which  determine  the  type  of  the  orbit  in  the  fol- 
lowing forms: 

I.     When  the  velocity  of  projection  equals  the  velocity 

from  infinity  the  orbit  is  a  parabola. 
II.     When  the  velocity  of  projection  is  less  than  the 

velocity  from  infinity  the  orbit  is  an  ellipse. 
III.     When  the  velocity  of  projection  is  greater  than  the 
velocity  from  infinity  the  orbit  is  a  hyperbola. 

Thus  if  a  comet  starts  from  rest  at  an  infinite  distance  from 
the  sun  and  falls  towards  the  sun  its  orbit  will  be  a  parabola. 
If  it  is  projected  towards  the  sun  from  an  infinite  distance 
its  orbit  will  be  a  hyperbola.  If  it  falls  from  rest,  starting 
from  a  finite  distance,  its  orbit  will  be  an  ellipse. 

223.   Period  of  Revolution.  —  From  equation  (III)  we  have 

h  =  r2w  =  r~  —  - 
dt 

:.     hdt=r  -rdd=2dA, 

where  dA  is  the  area  swept  over  by  the  radius  vector  in  the 
time  dt.  Therefore  when  the  orbit  is  an  ellipse  the  period  of 
revolution  is 

P  =  fPdt 
Jo 


O      n-nab 

-  I    dA      {irab  =  area  of  ellipse) 
h  Jq 


O      f*irab 
h 

2wab 


where  a  and  6  are  the  semi-major  axis  and  semi-minor  axis 
of  the  ellipse,  respectively.     But  by  equations  (6) 

h  =  Vep.M 
and  by  the  properties  of  the  ellipse  ep  =— ,  therefore  h  =  y- 


a  T  a 


MOTIOX  OF  A  PARTICLE  293 

Substituting  the  last  expression  for  h  in  that  for  P  we  obtain 

V/i 

2  ircA 


V7  (M  +  m) 

It  will  be  noted  that  the  period  of  revolution  depends  upon 
the  major  axis  but  not  upon  the  minor  axis  of  the  orbit. 

The  results  obtained  in  discussing  the  motion  of  two  gravi- 
tating  particles  are  as  they  appear  to  an  observer  who  is  lo- 
cated on  one  of  the  bodies.  The  form  and  size  of  the  orbit, 
the  period  of  revolution,  etc.,  will  be  the  same  whether  the 
observer  is  located  on  one  or  on  the  other  of  the  two  bodies. 
For  instance,  to  an  observer  on  the  moon  the  earth  describes 
an  orbit  which  is  exactly  similar  to  the  orbit  which  the  moon 
appears  to  describe  to  an  observer  on  the  earth. 

224.  Mass  of  a  Planet  which  has  a  Satellite.  —  In  order  to 
fix  our  ideas  let  the  earth  be  the  planet.  Then,  since  the  ac- 
celeration due  to  the  sun  is  practically  the  same  on  the  moon 
as  it  is  on  the  earth,  the  period  of  revolution  of  the  moon 
around  the  earth  is  the  same  as  if  they  were  not  in  the  gravi- 
tational field  of  the  sun.  Therefore  the  period  of  the  moon 
around  the  earth  is 

p,  2rt'l 


v7  (w  +  ttc') 
while  that  of  the  earth  around  the  sun  is 

-  -"- 


where  .1/",  ///,  and  ///'  arc  the  masses  of  the  sun.  of  the  earth, 
and  of  the  moon,  respectively,  a  is  t  he  semi-major  axis  of  the 
earth's  orbit,  and  a'  thai  of  the  moon's  orbit. 

Squaring  these  equation-  and  dividing  one  by  the  other 


±£Lm(L\\(*) 

+  ni     V/'V   w 


294  ANALYTICAL  MECHANICS 

Since  m'  is  negligible  compared  with  ra,  and  in  compared  with 
M,  the  last  equation  may  be  written  in  the  form 


m 
M 


©'•£)' 


which  gives  the  ratio  of  the  mass  of  the  planet  to  that  of  the 
sun. 

225.  Kepler's  Laws.  —  In  establishing  the  truth  of  the  law 
of  gravitation  Newton  showed  that  the  same  law  which 
makes  the  apple  fall  to  the  ground  keeps  the  moon  in  its 
orbit.  Then  he  extended  the  application  of  the  law  to  the 
other  members  of  the  solar  system  by  accounting  for  the 
empirical  laws  which  Kepler  (1571-1630)  had  formulated 
from  the  observations  of  Tycho  Brahe  (1546-1601).  The  fol- 
lowing are  the  usual  forms  in  which  Kepler's  laws  are  stated. 

1 .  Each  planet  describes  an  ellipse  in  which  the  sun  occu- 

pies one  focus. 

2.  The  radius  vector  describes  equal  areas  in  equal  inter- 

vals of  time. 

3.  The  square  of  the  period  of  any  planet  is  proportional 

to  the  cube  of  the  major  axis  of  its  orbit. 

The  first  law  is,  as  we  have  seen,  a  direct  consequence  of 
the  inverse  square  law. 

The  second  law  follows  from  equation  (III),  which  holds 
good  tor  ;ill  bodies  moving  in  central  fields  of  force. 

'I'lic  third  law  amounts  to  stating  that  the  masses  of  the 
plaints  are  negligible  compared  with  the  mass  of  the  sun. 
For  if  ///.  a,  and  /'  refer  to  one  planet  and  m',  a',  and  P'to 
another  planet,  then 

P=     ,    2*«'         andP'- 


V7  (M  +  m)  Vy  (M  +  m') 

Therefore 


try     taY  M  +m' 

'/'7       Uv  '  M  +  m' 


MOTION  OF  A  PARTICLE 
Evidently  when  m  and  m*  are  negligible  compared  with  M 

©■-6)'  • 

which  is  of  Kepler's  third  law. 


PROBLEMS. 

1.  The  gravitational  acceleration  at  the  surface  of  the  earth  is  about 
980  cm. /sec.2  Calculate  the  mass  and  the  average  density  of  the  earth, 
taking  6.4  X  108  cm.  for  the  mean  radius,  and  supposing  it  to  attract  as 
if  all  its  mass  were  concentrated  at  its  center. 

2.  The  periods  of  revolution  of  the  earth  and  of  the  moon  arc,  roughly, 
365^  and  27  J  days.  Find  the  mass  of  the  moon  in  tons.  Take  6.0  X  1027 
gm.  for  the  mass  of  the  earth. 

3.  The  periods  of  revolution  of  the  earth  and  of  the  moon  are  365^ 
and  27|  days,  respectively,  and  the  semi-major  axes  of  their  orbits  are, 
approximately,  9.5  X  107  and  2.4  X  105  miles.  Find  the  ratio  of  the  mass 
of  the  sup  to  that  of  the  earth. 

4.  Taking  the  period  of  the  moon  to  be  27£  days,  and  the  radius  of 
its  orbit  to  be  3. So  X  1010  cm.,  show  that  the  acceleration  of  the  moon. 
due  to  the  attraction  of  the  earth,  is  equal  to  what  would  be  expected 
from  the  gravitational  law.  Assume  the  gravitational  acceleration  at 
the  surface  of  the  earth,  that  is,  at  a  point  6.4  X  108  cm.  away  from  the 

center,  to  be  980  —  • 


sec 


6.  Show  that  if  the  earth  were  suddenly  stopped  in  its  orbit  it  would 
fall  into  the  sun  in  about  62.5  days. 

6.  Show  that  if  a  body  is  projected  from  the  earth  with  a  velocity  of 
7  miles  per  second  it  may  Leave  the  solar  system. 

GENEK  \l.    PROBLEMS. 

1.  Find  the  expression  for  the  central  force  under  which  a  particle 
describes  the  orbit  rn  =  an  cos  >td  and  consider  the  special  cases  when 

(a)  n  =  i,  (c)  »  =  1,  (c)  n  =  2. 

(b)  n=-h,  («1)  n  =  2, 

2.  A  particle  moves  in  a  central  field  of  force  with  a  velocity  which  is 
inversely  proportional  to  the  distance  from  the  center  of  the  field.  Show 
that  the  orbit  is  a  logarithmic  Bpiral. 


296  ANALYTICAL   MECHANICS 

3.  A  gun  can  project  a  shot  to  a  height  of  — ,  where  R  is  the  radius  of 
the  earth.  Taking  the  variation  of  the  gravitational  force  with  altitude, 
show  that  the  gun  can  command  —of  the  earth's  surface. 

4.  A  panicle  is  projected  into  a  smooth  horizontal  circular  groove. 
The  particle  is  attracted  towards  a  point  in  the  radius  which  joins  the 

position  nf  projection  with  the  center,  with  a  force  equal  to  ~  Show  that 
in  order  that  the  particle  may  be  able  to  make  complete  revolutions  the  ini- 
tial velocity  must  not  be  less  than       ^      ,  where  a  is  the  radius  of  the 

aL  —  b- 

groove  and  b  the  distance  of  the  center  of  force  from  the  center  of  the 
groove. 

5.  A  cornel  describing  a  parabolic  orbit  about  the  sun  collides  with  a 
body  of  equal  mass  at  rest.  Show  that  the  center  of  mass  of  the  two 
describes  a  circle  about  the  sun  as  center. 

6.  Prove  that  the  least  velocity  with  which  a  body  must  be  projected 
from  the  ninth  pole  so  as  to  hit  the  surface  of  the  earth  at  the  equator  is 
about  1}  miles  per  second,  and  that  the  angle  of  elevation  is  22°.o. 

7.  A  particle  moves  in  the  common  field  of  two  fixed  centers  of  force 
of  equal  intensity.  The  particle  is  attracted  towards  one  of  the  centers 
with  a  force  which  varies  as  its  distance  from  that  center,  and  repelled 
from  the  other  center  according  to  the  same  law.  Show  that  the  orbit 
is  a  parabola. 

8.  A  particle  moves  in  a  field  in  which  the  force  is  repulsive  and  varies 

ly  as  the  square  of  the  distance  from  the  center  of  force.     Show 
thai  the  orbil  is  a  hyperbola. 

9.  In  the  preceding  problem  show  that  the  radius  vector  sweeps  over 
equal  areas  iu  equal  intervals  of  time. 


CHAPTER   XV. 
PERIODIC    MOTION. 

226.  Simple  Harmonic  Motion.  —  When  a  particle  moves  in 
a  straight  line  under  the  action  of  a  force  which  is  directed 
towards  a  fixed  point  and  the  magnitude  of  which  varies 
directly  as  the  distance  of  the  particle  from  the  fixed  point, 
the  motion  is  said  to  be  simple  harmonic. 

Let  0,  Fig.  130,  be  the  fixed  point,  m,  the  mass  of  the  par- 
ticle, and  x  its  distance  from  0;  then  the  foregoing  definition 
gives 

F=-kx,  (I') 


O  mX.      m 

Fig.  130. 

where  k  is  the  constant  of  proportionality.  The  negative 
sign  in  the  right-hand  member  of  the  equation  (I')  accounts 
for  the  fact  that  F  is  directed  towards  the  fixed  point,  while 
x  is  measured  in  the  opposite  direction.  Substituting  this 
expression  for  F  in  the  force  equation  we  get 

mjt=  -kx,  (I) 

I---*  (I"> 

where  w2=—.     Substituting?'-      U>r       inequation  (I")  and 
m  dx         dt 

integrating  we  have 

y2  =  c2  —  co'-'.r'-. 

Let  v=  Va  when  x  =  0,  then  c  =  r0.    Therefore 


v  =  VVo2-o>2x2.  (II) 

297 


298  ANALYTICAL  MECHANICS 

Putting  equation  (II)  in  the  form 


dx  t'o1 


■■2 


and  integrating  we  obtain 

•     ,  wo: 
sin-1—  =  ut+8, 

Vn 

or  x  = -sin  (<at+ 8) 

CO 

=  asin  (ut+8),  (III) 

where  5  is  the  constant  of  integration  and  a  =  —  • 

CO 

227.  Displacement.  —  The  distance,  x,  of  the  particle  from 
the  fixed  point  is  called  the  displacement. 

228.  Amplitude.  —  The  maximum  displacement  is  called 
the  amplitude.  It  is  evident  from  equation  (III)  that  the 
amplitude  equals  a. 

229.  Phase.  —  The  particle  is  said  to  be  in  the  same  phase 
at  two  different  instants,  if  the  displacement  and  the  velocity 
at  the  «>ne  instant  equal,  respectively,  the  displacement  and 
the  velocity  at  the  other  instant. 

230.  Period.  —  The  time  which  elapses  between  two  suc- 
cessive instants  at  which  the  particle  is  in  the  same  phase  is 
called  the  period  of  the  motion.  In  order  to  find  the  period 
we  will  make  use  of  the  definition  of  a  periodic  function.* 
It  is  evident  from  equation  (III)  that  x  is  a  periodic  function 
of  t:  therefore  we  can  write 

x  =  a  sin  [cot  +  <5] 
=  asin[«(*+P)  +  «]. 

iy  variable  x  is  :i  periodic  function  of  any  other  variable  t  and  if  the 
dependence  <if  z  on  /  is  given  by  the  elation  x  =  /  (t),  then  the  function  satis- 
fiea  the  following  condition: 

/(0-/(H 
•  period  and  »  any  positive  or  negative  integer.    As  an  illustra- 
tbe  function  x  =  sin  9.     This  function  evidently  satisfies  the 
Therefore  2  w  is  the  period. 


PERIODIC   MOTION 


299 


But  since  sin  0  is  a  periodic  function  of  6  witji  a  period  of  2  tt, 
we  have 

sin  0  =  sin  (d+2ir), 

t  herefore  x=  a  sin  [«  (<  +  P)  +  5  ] 

=  a  sin  M  +  5  +  2  tt] 

2tt 


and  consequently 


P  = 


(IV) 


231.  Frequency.  —  The  number  of  complete  vibrations 
which  the  particle  makes  per  second  is  called  the  frequency 
of  the  vibration.     If  n  denotes  the  frequency,  then 


U=P 


(V) 


232.    Time-distance   Diagram.  —  Suppose   the   particle    to 
describe  the  vertical  line  AA',  Fig.  131,  the  middle  point  of 


Fig.   L31. 


which  is  the  fixed  point.  Then  OA  =  OA'  =  a  =  the  ampli- 
tude. The  relation  between  the  position  of  the  particle  and 
the  time  may  be  visualized  by  plotting  equation  (III)  with 
x  as  ordinate  and  t  as  abscissa.  This  gives  the  well-known 
sine  curve. 

A  mental  picture  of  the  motion  of  the  particle  may  be 
formed  by  supposing  thai  the  particle  under  consideration 
is  a  projection  of  another  particle  which  moves  in  a  circle 
of  radius  a  with  a  constanl  speed.  1 1 1 * ■  second  particle  and 
its  path  may  be  called  the  auxiliary  particle  and  the  auxiliary 
circle,  respectively. 


300 


ANALYTICAL   MECHANICS 


(HI') 


233.  Common  Forms  of  Equation  (III).  —  The  following  are 
the  typical  forms  in  which  equation  (III)  is  written: 

x  =  a  sin  (wt  +  S)  =  a  sin  w  (/  +  fo) 

=  a  sin  [-£t+  6]  =  a  sin  -—  (t  +  to) 

=  a  cos  (wt  +  5')  =  a  cos  w  (£  +  i/) 
=  acosf— ^+  8')  =  acos-^(£+^'), 

where  5'  =  -  —  8  and   fo'  —  - —  t0. 
2  4 

234.  Epoch.  —  The  constants  t0  and  k'  are  called  epochs, 
and  o  and  8'  are  called  epoc/i  angles.  The  meanings  of  these 
constants  will  be  seen  from  Fig.  131. 

235.  Velocity.  —  The  following  expressions  for  the  velocity 
of  the  particle  may  be  obtained  either  from  equation  (II)  or 
from  equation  (III) : 

v  =  Vv02  —  oi2x-  =  u  Va2  —  x2 
=  acocos  (cot+  8) 

=  aaj  sin(c^+  <5  + 


(I)  is  tho  displacement  -  time  curve 

(II)  is  the  velocity -time  curve 

Fro.  L32. 

I'   Lb  evident   from  these  expressions  that  the  velocity  is  a 
Bimple  harmonic  function  of  the  time,  that  it  has  the  same 


PERIODIC    MOTION 


301 


period  as  the  displacement,  and  thai  it  differs  in  phase  Prom 

P 
the  latter  by     ,  as  shown  in  Fig.  132. 
4 


236.    Energy  of  the  Particle, 
further  explanation. 


The  following  do  not  need 


-X' 


Fdx 


2  w-m 


2  iv-m    , 

Of. 

P2      X 

2Tr2a-m 
P2 


sin2  ^(t  +  to).  (VII) 


e=t+v 


p 

2ir-a2m 
P2 


2ir-a-m 


(a2  -  x2) 


P2 


(*+4).(vi) 


(VIII) 


Thus  the  total  energy  of  the  particle  is  constant  and  equals 
the  maximum  values  of  the  potential  and  kinetic  energies. 
The  total  energy  varies,  evidently,  directly  as  the  square  of 
the  amplitude  and  inversely  as  the  square  of  the  period. 

In  Fig.  133,  T,  U,  and  V  are  plotted  as  ordinates  and  the 
time  as  abscissa,  with  phase  relations  which  correspond  to 
the  curves  of  Fig.  132. 


(I)  is  the  Uand  t  Curve. 

(II)  is  the  T  and  t  Curve. 
(Ill)isthe  E and  t Curve. 

Fig.  133. 


237.  Average  Value  of  the  Potential  Energy.  Since  I '  may 
be  considered  as  a  function  of  either  x  or  t,  we  will  find  its 
average  value  with  respect  to  both  variables.     TakingO  and 


302  ANALYTICAL  MECHANICS 

a  as  the  limits  of  x  and  the  corresponding  values  of  t  as  the 

limits  of  t  we  have 

p  1        Ca 

i       /•>  U*= /    Udx 

77<=_i_j    Udt*  a-OJo 

f-o°  =-JrPx*dx 

4  p  2  a  Jo 

=  ±-"'"'  firing*  _^r 

P        P2      Jo  P  -     6 

...      E=sux=2  Ut-  (IX) 


PROBLEMS. 

1.  A  particle  which  describes  a  simple  harmonic  motion  has  a  period 
of ."»  Bee.  and  an  amplitude  of  30  cm.  Find  its  maximum  velocity  and  its 
maximum  acceleration. 

2.  When  a  load  of  mass  m  is  suspended  from  a  helical  spring  of  length  L 
and  of  negligible  mass  an  extension  equal  to  D  is  produced.  The  load  is 
pulled  down  through  a  distance  a  from  its  position  of  equilibrium  and  then 
sel  free.  Find  the  period  and  the  amplitude  of  the  vibration.  Hooke's 
law  holds  true. 

3.  Within  the  earth  the  gravitational  attraction  varies  as  the  distance 
from  the  center.  Suppose  there  were  a  straighl  shaft  from  pole  to  pole, 
with  no  resisting  medium  in  it.  What  would  be  the  period  of  oscillation 
of  a  body  dropped  into  the  shaft'.'  Suppose  the  earth  to  be  a  sphere  with 
a  radius  of  WOO  miles. 

4.  In  the  preceding  problem  find  the  velocity  with  which  the  body 
would  pass  the  center  of  the  earth. 

5.  A  particle  describes  a  circle  with  constant  speed.  Show  that  the 
projection  of  the  particle  upon  a  straight  line  describes  a  simple  harmonic 

motion. 

6.  The  pan  of  a  helical  spring  balance  is  lowered  2  inches  when  a 
weighl  of  ."»  pounds  is  placed  on  it.  Find  the  period  of  vibration  of  the 
balance  with  the  weighl  on. 

■  footnote  p.  L42. 


PEB  IODIC  MOTION  303 

7.  A  particle  which  is  constrained  to  move  in  a  straight  line  is  at- 
tracted by  another  particle  fixed  at  a  point  outside  the  line.  Show  that 
the  motion  of  the  particle  is  simple  harmonic  when  the  force  varies  as  the 
distance  between  the  particles. 

8.  A  particle  of  mass  m  describes  a  motion  defined  by  the  equation 

x  =  a  sin  (ojt  +  5). 

Find  the  average  value  of  the  following  quantities,  with  respect  to  the 

time,  for  an  interval  of  half  the  period: 

(a)  displacement;  (e)  momentum; 

(b)  velocity;  (f)  kinetic  energy; 

(c)  acceleration;  (g)  potential  energy. 

(d)  force; 

9.     In  problem  S  take  the  averages  with  respect  to  position. 

10.  In  problem  8  suppose  the  motion  to  be  given  by 

x  =  a  cos  oj  (t  +  t0). 

11.  In  problem  10  take  the  averages  with  respect  to  position. 

12.  In  problem  S  suppose  the  motion  to  be  given  by  the  following 
equations: 

I.  x  =  asm2  (cot +  8). 
II.  x  =  a  cos2  (ojt  +  5). 
III.    x  =  a  sin  iot  cos  (cct  +  5). 

238.  Composition  of  Two  Parallel  Simple  Harmonic  Motions 
of  Equal  Period.     Analytical  Method.    -  Suppose 

.r,  =  «i  sin  (<at  +  5X),  (1) 

and  x-i  =  a-i  sin  (at  +  52), 

to  define  the  motions  which  a  particle  would  have  if 
acted  upon,  separately,  by  two  simple  harmonic  forces. 
Then  the  motion  which  will  result  when  the  turn-  ad 
simultaneously  is  obtained  by  adding  equations  (1)  and  (2). 

Thus 

X  =  X\  + 

=  Oi  sin  (ojt  +  Si)  -f  Ch  sin  (at  +  5>). 


304 


ANALYTICAL   MECHANICS 


Expanding  the  right-hand  member  of  the  last  equation  and 
rearranging  the  terms  we  get 

x  =  (ai  cos  5i  +  a2  cos  52)  sin  wt 

+  (ai  sin  5i  +  «2  sin  S2)  cos  ut 
=  a  cos  S  sin  a>t  +  a  sin  5  cos  ut 
=  a  sin  («£  +  8), 
where         a  cos  8=  ai  cos  Si  +  a2  cos  82, 
and  a  sin  5  =  ai  sin  Si  +  a2  sin  82. 


(3) 


Fig.  134. 

It  is  evident  from  equation  (3)  that  the  resulting  motion  is 
simple  harmonic  and  has  the  same  period  as  the  component 
motion-. 

Squaring  the  last  two  equations  and  adding  we  obtain 
the  amplitude  of  the  motion  in  terms  of  the  constants  of 
equations  (1)  and  (2).     Thus 

a-  =  ar  +  o22  +  2  axa2  cos  (52  -  Si).  (4) 

The  phase  angle  of  the  motion  is  evidently  defined  by 

O]  sin  8\  +  a2  sin  52 


tan  6 


ai  cos  Si  +  a2  cos  S2 


(5) 


239.    Graphical  Method.  — The  graph  of  the  resulting  mo- 
tion may  be  obtained  by  either  of  the  following  methods: 

I     Represent    the  given  motions  by  displacement-time 
curves,  bhen  add  the  ordinates  of  these  curves  in  order  to 


PERIODIC  MOTION 


305 


obtain  the  curve  which  represents  the  resultant  motion.  In 
Fig.  135  the  curves  (I)  and  (II)  represent  the  component 
motions  and  curve  (III)  represents  the  resultant  motion. 


Fig.  135. 


(2)  Draw  two  concentric  auxiliary  circles  with  radii  equal 
to  the  amplitudes  of  the  component  motions;  draw  a  radius 
in  each  circle  making  an  angle  with  the  2-axis,  equal  to  the 
phase  angle  of  the  corresponding  motion;  the  vector  sum 
of  these  radii  gives  the  radius  of  the  auxiliary  circle  for  the 
resultant  motion  and  the  corresponding  phase  angle.  By 
the  help  of  this  auxiliary  circle  the  displacement-time  curve 
of  the  resulting  motion  can  be  drawn  without  drawing  those 
of  the  component  motions. 

PROBLEMS. 

Find  the  resultant  motion  due  to  the  superposition  of  two  motions 
denned  by  the  following  pairs  of  equations: 


(1)  Ji  =  ai  sin  ut  and  x2  =  flj.sin  f  wt  +  -]■ 

(2)  Xi  =  «i  sin  (at  and  Xj  =  '/..cos  (cot  —  -J- 

i  =  ai  cos  ut  and  x%  =  OjCOS  (ut  +  -  )• 

(4)  X\  —  Oi  sin  <at  and         =  Ojsin  (ut-\-8). 

(5)  X\  =  ax  sin  ut  and  x<  =  Otcoawt. 


306  ANALYTICAL  MECHANICS 

(6)  Xi  =  d  sin  -=■  t  and  x2  =  a2  cos  —  (t  +  t0). 

(7)  Xi  =  oi  cos  w<    and    x»  =  a*  eon  (cot  +  5). 

(8)  Xi  =  ai  cos  oj<    and    x»  =  a2  cos  (wi  +  5). 

(9)  xi  =  a,  cos  a'/     and     x2  =  a2cos  —  (t  +  t0). 

(10)  Xi  =  ai  sinfo}«  +  |j     and  a*  =  flfesin  —  t. 

(11)  xi  =  <h  sin  (a;/  +  50      and   x2  =  «2  cos  (cot  +  52). 

(12)  xi  =  «i  cos  (ut  +  5,)       and   x2  =  a2  cos  (u*  +  S2). 

(13)  Ji  =  ai  cos  -£■  (t  +  t„)   and   x2  =  o^sin^  (/  -  t0). 
Mi    Xi  =  a,  sin  ^~(t-  Q    and   x2  =  a2  sin y  (<  +  f0). 

(15)   xi  =  ai  cos  -£(t—  'o)    and  x2  =  a2  cos  —  (t  +  /0)- 

240.  Elliptic  Harmonic  Motion.  —  Consider  the  motion  of  a 
particle  which  is  acted  upon  by  two  harmonic  forces  whose 
directions  are  perpendicular  to  each  other.  Suppose  the 
periods  of  vibration  of  the  particle  due  to  the  separate  action 
of  the  forces  to  be  the  same,  then  the  following  equations 
define  the  component  motions. 

x  =  a  sin  cot*  (1) 

y=bsm(at  +  8).  (2) 

The  equation  of  the  path  of  the  particle  may  be  obtained 
by  eliminating  /  between  equations  (1)  and  (2).  Expanding 
the  right-hand  member  of  equation  (2)  and  substituting  for 
sin  tat  and  cos  tat  from  equation  (1)  we  get 

y  =  b  sin  cot  •  cos  5+6  cos  cot  •  sin  5 

=  b-  cos  5  +  6  V  1  -  ^-  sin  5, 
a  .     *         a- 

'  The  phase  angle  is  left  out  of  equation  (1)  to  simplify  the  problem.    This, 
however,  does  nol  affecl  the  generality  of  the  problem.     It  simply  amounts  to 
ng  a  particular  instanl  as  the  origin  of  the  time  axis.     If,  however,  the 
ogle  i-  left  oul  of  l»>tli  :>f  the  component  motions  the  generality  <>f  the 
problem  it  affected  because  thai  will  amount  to  assuming  thai  the  compo- 
nent motions  .-ire  in  the  aame  phase. 


or 


y     x 

?  —  cos 

0 


PERIODIC  MOTION 


307 


•  sin  5. 


a  *  a- 

Squaring  the  last  equation  and  simplifying  we  have 

x2  ,  y2      2  xy        .       .   ., 

— ;  +  To r  cos  8  =  sin2  8, 

o2      o-       ab 

which  is  the  equation  of  an  ellipse,  Fig. 
136.  The  following  cases  are  of  special 
interest. 

Case  I.     When  5  =  0,  equation  (3)  re- 
duces to  y  =  -  x,  which  is  the  equation  of 

the  line  AA'.  Substituting  the  values 
of  x  and  y  in  the  equation 


(3) 


r  =  v  x2  +  y2 
we  obtain 

r  =  Va2  +  b2  •  sin  ut  FlG-  136- 

for  the  equation  of  the  motion.     Therefore  the  motion  is 
simple  harmonic,  in  the  line  AA' ,  with  an  amplitude  equal 

to  v / a2  -f  b2  and  period  —  • 

Case  II.     When  8  =  v,  equation  (3)  reduces  to  y  =  — x. 

a 

Therefore  the  motion  is  similar  to  that  in  Case  I  and  takes 

place  in  the  line  BB'. 

■Case  III.     When  8  =  ±  ~  equation  (3)  reduces  to 
*L2  +  ^2  =  1 


while  equations  (1)  and  (2)  become 

x  =  a  sin  cot, 
y  =  b  cos  ut. 

In  this  case,  therefore,  fche  particle  describes  an  elliptical 


308  ANALYTICAL  MECHANICS 

path  with  a  period  equal  to  —  .    The  axes  of  the  path  coin- 

co 

cide  with  the  coordinate  axes. 

Case  IV.  When  S  =  =fc  -  and  b  =  a,  the  path  becomes  a  cir- 
cle, and  the  motion  uniform  circular  motion  with  a  period 
equal  to—  • 

u> 

PROBLEMS. 

Find  the  resultant  motion  due  to  the  superposition  of  the  motions 
defined  by  the  following  equations: 

(1)  s  =  a  cos  cot  and  y  =  a  smut. 

(2)  x  =  a  cos  —  (t  +  /0)   and  y  =  a  sin  —  t. 

(3)  x  =  asm  cot  and   y  =  a  cos  —  (/  +  t0). 

(4)  x  =  a  sin  (cot  +  5)   and  y  =  a  cos  (co£  —  5) . 

(5)  x  =  a  cos  co£   and   y  =  b  sin  a>£. 

(6)  x  =  a  sin  (cot  —  5)   and    (/  =  b  cos  (oj£  +  8). 

(7)  x  =  a  sin  (cot  —  -J  and   ?/  =  6  sin  { cot  +  -)• 

(8)  x  =  acosf  w/ +  M  and   y  =  6sin  (lot  —  -}■ 

•(-9 

(10)   x  =  a  cos  M  +  6i)  and  ?/  =  6  sin  (co£  +  52). 

241.  Physical  Pendulum.  —  Any  rigid  body  which  is  free  to 
oscillate  under  the  ad  ion  of  its  own  weight  is  called  a  physical 
or  a  compound  pendulum.  Let  A,  Fig.  137,  be  a  rigid  body 
which  is  tree  to  oscillate  about  a  horizontal  axis  through  the 
point  0  and  perpendicular  to  the  plane  of  the  paper.  Fur- 
ther  lo  c  denote  the  position  of  the  center  of  mass  and  D  its 
distance  from  the  axis.     Then  the  torque  equation  gives 

/'^  =  -mgDBmd,  (X) 

where  m  is  the  mass  of  the  body  and  0  the  angular  displace- 
ment from  the  position  of  equilibrium. 


(9)    x  =  a  cos  [cot  —   J  and   y  =  b  cos  (  cot  +  - 


PERIODIC  MOTION 


309 


The  equation  — :T  =  -k-sinx  is 
ar 

not  integrable  in  a  finite  number 

of  terms;  therefore  the  solution  of 

equation  (X)  must  be  given  either 

in  an  approximate  form,  or  it  must 
be  expressed  as  an  infinite  series. 
First  Approximation.  —  When 
6  is  small  sin  0  may  be  replaced  by 
6.  Therefore  we  can  write 
rdco 


dt 


mgDd, 


(X') 


Fig.  137. 


dt 


(X") 


where 


mgD 
I    ' 


It 


be  observed  that   the  last    two 


equations  are  of  the  same  type  as  equations  (I')  and  (II')  of 
p.  297,  the  differential  equations  of  simple  harmonic  motion. 
Therefore  the  motion  of  the  physical  pendulum  is  approxi- 
mately harmonic.  Hence  we  can  apply  to  the  present  prob- 
lem the  results  which  were  obtained  in  discussing  simple 
harmonic  motion.     Thus  the  expression  for  the  displace- 


ment i^ 


a  sin  (ut+  8), 


XI) 


where  a  is  the  amplitude,  i.e.,  the  maximum  angular  dis- 
placement of  the  pendulum.  On  the  other  hand  the  period 
of  the  pendulum  is  2  v 

^o  =  — r 


=  2ttv/ 


\  mgD 


/Ir+mD* 

-  ""  V        — ~ — 
>       mgD 


=  2 


IK-+D2 


XII 


(XIII) 


310  ANALYTICAL  MECHANICS 

where  /  is  the  momenl  of  inertia  of  the  pendulum  about  an 
axis  through  the  center  of  mass  parallel  to  the  axis  of  vibra- 
tion and  K  is  the  corresponding  radius  of  gyration. 

Second  Approximation.  —  Starting  with  the  energy  equa- 
tion we  have 
'ddV 


x(D! 


mgh 

=  mgD  (cos  6  —  cos  a), 
or        dt  = 


I                 dd 

2  mgD    VCos  0  -  cos  a 
r               dd 

V  mgD      1  .     a       .   ,  e 
y/sin^-sm'- 

cos0=  1  -  2  sin2- 

Integrating  the  left-hand  member  of  the  last  equation  be- 

turn i  the  limits  t  =  0  and  t=  -.  and  indicating  the  integra- 

4 

t  ion  of  the  right-hand  member  between  the  corresponding 

limits  we  have 


P_ 

4~ 


Jo    \/sin"2"Sm 


The  last  integral  cannot  be  evaluated  in  a  finite  number  of 
terms,  but  we  can  expand  the  integrand  into  a  power  series, 
every  term  of  which  is  integrable. 

Lei  sin  -  =  sin- sin  0. 

Then  <;>  -  0  when  6=  0,  and  </>=  -when  0=  a;  further 

2  sin  -  cos  0  •  d<p 
dd=-    -±—  -• 

y  l-sin2|sin-</> 


PERIODIC  MOTION 


311 


Making  these  substitutions  in  the  left-hand  member  of  the 

preceding  expression  for  the  period  we  obtain 


P=4 


\   mgD 


(iDJa    V  " 


sin-"  sin2  <f> 


'd<t>* 


;  + 

4\/-^-  /'Yl+isin^sinV  +  -1^-sin^sin^  +  -  •  -W 
V  mgDJo  \       2        2  1-2-3        2  / 


V  mgDJo  L 


l  +  ^sin2^(l-cos2  0)  + 


]ch 


=4i/5(2 


4y 


.    7r    .    ,  a   , 


) 


i      p     1       ■      »ff     i 

1  +  iSin"2  + 


1  +  ^sin- 


•)       [by  (XII)] 

["when  a  is  small  higher "1 
L  terms  may  be  neglected  J 


$     [* 


hen  a  is  small  sin 


Therefore 

242.  Simple  Pendulum.  —  A  ball 
which  is  suspended  by  means  of  a 
string  forms  a  simple  pendulum  when 
it  is  free  to  swing  about  a  horizontal  / 

axis   through    the  upper  end  of  the  / 

string,  provided  the  mass  of  the  string 
is  negligible  compared  with  that  of  the     / 
ball  and  the  radius  of  the  ball  is  negli-  v. 
gible  compared  with  the  length  of  the 
string.     If  m  denotes  the  mass  of  the 
ball  and  I  the  length  of  the  string  then 

*  This  is  called  an  elliptic  integral. 

t  This  expansion  1-  carried  nut  by  the  Binomial  theorem. 
X  See  Appendix  At. 


I 


(XIV) 


Flo.  138 


See  Appendix  A. 


312  ANALYTICAL   MECHANICS 

the  moment  of  inertia  of  the  pendulum  equals  ml2.  Therefore 
substituting  this  value  of  /  in  the  expressions  for  P0  and  P 
and  replacing  D  by  /  we  obtain 

Po=27rV/-^- 

y  mgD 

=  2irl/-'  (XV) 

*  9 

for  the  first  approximation,  and 

=  P.(l  +  $  (XV) 

for  the  second  approximation. 

243.  Equivalent  Simple  Pendulum.  —  A  simple  pendulum 
which  has  the  same  period  as  a  physical  pendulum  is  called 
the  equivalent  simple  pendulum  of  the  latter.  If  I  denotes  the 
Length  of  the  equivalent  simple  pendulum,  then 

>  g  y     gD 

.-.     i  =  R2  +  D\  (XVI) 

For  a  giveo  value  of  D  and  a  given  direction  of  the  axis,  K 
is  constant.  Therefore  if  the  direction  of  the  axis  is  not 
changed  I  is  a  function  of  I)  alone.  If  we  plot  the  last  equa- 
tion with  /  as  ordinate  and  D  as  abscissa  we  obtain  a  curve 
similar  to  that  of  Fig.  139.  It  is  evident  from  the  curve  that 
the  value  of  I  is  infinitely  large  for  D=  0,  but  it  diminishes 

2  K 
rapidly  to  the  minimum  value——  as  D  reaches  the  value  A'. 

9 

\-  />i-  increased  further/  increases  continually.    It  will  be  ob- 

2  A 
served  that  for  a  given  value  <»f  /  greater  than"     there  are  two 

9 
values  of  D,  one  of  which  is  less  and  the  other  greater  than  K. 


PERIODIC    MOTION 


313 


The  group  of  parallel  axes  about  which  the  rigid  body 
oscillates  with  the  same  period  forms  two  coaxial  circular  cyl- 
inders,. Fig.  140,  whose  common  axis  passes  through  the  cen- 


D,->i 


Fig.  140. 

ter  of  mass.    The  cylinders  which  correspond  to  the  minimum 
value  of  the  period  coincide  and  have  a  common  radius  K. 

PROBLEMS. 

1.  Find  the  period  of  the  following  physical  pendulums: 

(a)  A  uniform  rod,  i lie  transverse  dimensions  of  which  arc  negligible 
compared  with  the  length,  oscillates  aboul  a  horizontal  axis  through  one 
end. 

(b)  A  sphere  suspended  from  a  horizontal  axis  by  means  of  ;i  string  of 
negligible  mass.  Discuss  the  changes  in  the  period  as  the  axis  approaches 
the  center  of  the  sphere. 

(c)  A  circular  flat  ring  oscillates  aboul  an  axis  which  forms  an  element 
of  the  inner  surface. 

\  door  oscillates  aboul   the  line  of  the  hinu.es  which  make  an  angle 
ex  with  the  vertical. 

2.  A  sphere  of  radius  a  oscillates  hack  and  forth  in  a  perfectly  smooth 
spherical  howl  of  radius  >>.  Find  the  period  of  oscillation.  The  sphere 
i.>  supposed  to  have  no  rolling  motion. 

3.  'What  effect  on  the  period  of  a  pendulum  would  he  produced  by  a 
change  in  the  mass  of  the  bob,  or  of  the  length  of  the  string,  or  in  the 

radius  of  the  earth,  or  in  the  length  of  the  da\  ,  or  in  the  latitude  of  the 
location? 

4.  A  seconds  pendulum  lo8es  -'ID  seconds  per  day  at  the  summit  of  ;i 
mountain.     Find  the  heighl  of  the  mountain,  considering  the  earth  to  he 


314  ANALYTICAL  MECHANK  S 

a  sphere  of  4000  miles  radius  and  the  gravitational  force  to  vary  inversely 
as  the  square  of  the  distance  from  the  center  of  the  earth. 

5.  Given  the  heighl  of  a  mountain  above  the  surrounding  plain  and 
the  period  of  a  pendulum  on  the  plain  and  on  the  top  of  the  mountain, 
find  a  relation  from  which  the  radius  of  the  earth  can  be  computed. 

6.  Supposing  the  gravitational  attraction  within  the  earth  to  vary  as 
the  distance  from  the  center,  find  the  depth  below  the  surface  at  which  a 
seconds  pendulum  will  beat  2  seconds. 

7.  Derive  a  relation  between  the  distance  of  a  pendulum  from  the 
cuter  of  the  earth  and  its  period. 

8.  A  balloon  ascends  with  a  constant  acceleration  and  reaches  400  feet 
in  one  minute.  What  is  the  rate  at  which  the  pendulum  gains  in  the  bal- 
loon? 

9.  A  pendulum  of  length  I  is  shortened  by  a  small  amount  5/.     Show 

that  it  will  gain  about  ^r- vibrations  in  an  interval  of  time  of  n  vibra- 
tions,    n  is  supposed  to  be  a  large  integral  number. 

10.  How  high  above  the  surface  of  the  earth  must  a  seconds  pendu- 
lum be  carried  in  order  that  it  may  have  a  period  of  4  seconds? 

11.  While  a  train  is  taking  a  curve  at  the  rate  of  60  miles  per  hour  a 
seconds  pendulum  hanging  in  the  train  is  observed  to  swing  at  the  rate  of 
L21  oscillal  ions  in  2  minutes.  Show  that  the  radius  of  the  curve  is  about 
a  quarter  of  a  mile. 

12.  bind  the  expressions  for  the  least  period  of  oscillation  the  following 
bodies  can  have;  also  determine  the  corresponding  position  of  the  axes. 

(a)  Rod  of  negligible  transverse  dimensions.         (d)    Solid  cylinder. 

(b)  Square  plate  of  negligible  thickness.  (e)    Solid  sphere. 

(c)  Circular  plate  of  negligible  thickness.  (f)    Spherical  shell. 

244.  Determination  of  the  Gravitational  Acceleration  by 
Means  of  a  Reversible  Pendulum.  —  A  physical  pendulum 
which  is  provided  with  two  convenient  axes  of  vibration  is 
called  a  reversible  pendulum.  Let  Dand  D',  Fig.  141,  denote 
the  distances  of  the  axes  from  the  center  of  mass.  Then  the 
corresponding  periods  are 

>        gD 


PERIODIC   MOTION 


:;i:> 


Eliminating  K  and  solving  for  g  we  get 

°-*r'D>P>*-DF>'  (1) 

Eleversible  pendulums  which  are  made  for  the  purpose  of 
determining  g  are  so  constructed  that  the  two  periods  are 
very  nearly  equal.     Therefore  we  can  write 

P'=P  +  8P,     [8P^P], 
and  obtain 

,     ,  D'--D2 


D'  (P 


=  4tt- 


=    4  7T: 


=  4 


spy- -dp2 

D'2  -  D2 


P2  {D'-D)  +  2  PD'SP  +  Z)'  (5P)2 
D  +D' 


2D' 


1/ 


[(8p)2  is  neglected] 


1  - 


D    PJ 

2D' 

D'-D 


f) 


(2) 


The  approximate  expression  which 
is  given  in  equation  (2)  is  better 
adapted  for  computing  the  value  of 
g  from  experimental  data  than  the 
more  exact  expression  given  in 
equation  (1).  This  is  due  to  the 
fact  that  (D'  —  D),  which  cannot  be 
determined  with  a  high  degree  of 
accuracy,  enters  into  equation  ( 1) 
as  a  factor,  while  it  appears  only  in 
the  correction  term  of  equation  (2). 
245.  Bifilar  Pendulum.  —  A  rigid 
body  which  is  suspended  by  means 
of  two  parallel  strings,  as  shown  in 
Fig.  142,  is  called  a  bifilar  pendulum 


A 


x 


D 


/ 


li...   1  11. 

When  the  body  is 

given  an  angular  displacement  aboul  a  vertical  axis  through 


See  Appendix  Ai. 


316 


ANALYTICAL  MECHANICS 


.... 

«-2D-» 

h 

1   » 

a 

h 

1  1   1  ' 

its  center  of  mass  and  then  left  to  itself  it  vibrates  with  a 
definite  period. 
Let      m  =  the  mass  of  the  rigid  body, 

/  =  the  moment  of  inertia  of  the 

body  about  a  vertical  axis 

through  its  center  of  mass, 

I  =  the  length  of  each  string, 

0  =  the   angular   displacement   of 

the  bar, 
<j>  =  the  angular  displacement   of 

the  strings, 
T  =  the  1  ensile  forces  of  the  strings, 
and 
2  D=  the     distance     between     the 

si  rings.  1  | 

,      •      ,  •  FlG-  142- 

In  order  to  obtain  the  torque  equation 

suppose  the  weight  of  the  suspended  system  to  be  concen- 
trated at  the  ends  of  the  small  bar  ab  and  analyze  the  forces 
acting  upon  it  as  shown  in  Fig.  143.  Evidently  ab  is  acted 
upon  by  four  forces,  namely,  the  tensile  forces  of  the  strings 
and  the  two  forces  each  of  which  represents  half  the  weight 
of  the  suspended  system.  These  forces  are  equivalent  to  a 
couple  formed  by  the  forces  Fand  —  F,  which  act  at  the  ends 
of  ab  in  a  horizontal  direction,  and  a  vertical  force  equal  to 
the  difference  between  the  sum  of  the  vertical  components 
of  the  tensile  forces  of  the  strings  and  the  weight  of  the 
suspended  Bystem.  The  vertical  force  gives  the  suspended 
n  a  motion  in  the  vertical  direction.  But  both  this 
motion  and  the  force  which  produces  it  are  very  small 
therefore  they  will  he  neglected. 

It    i-  evident    from  Fig.  143  that    the   torque  due  to  the 
horizontal  couple  is 


Q 


2.  F-l)  cos 


im:i;i<>i>[<'  motion 


317 


Substituting  this  value  of  G  in  the  torque  equation  we  hav 


I 


dt 


-2FDcos- 


=  -  2  TD  sin  ^  cos  - . 


First  Approximation.  —  When  0  and  0  are  small  the  fol- 
lowing relations  give  close  enough  ap- 
proximations. 

T  =  h  mg, '   cos-  =  1, 


—  I'D 


Dd  =  1(f)*    sin  0  =  0 


?<■ 


flaking  these  substitutions  in  the  torque 
equation  we  get 

Tdu         mgD'2 


dt 


which  is  the  equation  of  simple  harmonic 
motion.     Therefore 


»=4?  V- 

D    »  mg 


is  the  period  of  the  motion. 

Second  Aimm;<  >\i.\iation.  —  From  Fig. 
143  we  have 


T  cos  <f>=%mg    and     ea'  =  lsin<t>  =  2  Dsin-  • 


Therefore 
and 


.  2D   .    0 

sin  0  =  —  sin  - 


T  = 


mg 


mg 


'_'  COS  0 


n/-4-;'-", 


*  The  lineea'  is  considered  as  an  arc  of  each  ol  two  circles  with  centers  a1  g  and  c 


318 


ANALYTICAL  MECHANICS 


Making  these  substitutions  in  the  torque  equation  we  obtain 


rfco  _      nujl)- 
dt~  I 


rm/I)- 


2  sin -cos- 


i- 


4  D- 


sm 


\ 


4  d-  .  .,  e 


Iii  actual  experiments       is  made  less  than  —  •     Therefore 

even  if  the  maximum  value  of  0  is  made  as  large  as  half  a 
radian  the  second  term  under  the  radical  is  less  t  haii  ,  f^  and 
consequently  negligible.     Thus  the  last  equation  reduces  to 

du>  _     mgD*. 

which  is  the  well-known  pendulum  equation.  Therefore  we 
have 


D  V    mg  V 
16/ 


1  + 


L6 


=  n  i 


for  a  second  approximation  to  the  actual 
value  of  I  hf  period. 

246.  Torsional  Pendulum.  —A  torsional 
pendulum  consists  of  a  rigid  bodysuspended 
by  a  wire,  the  wire  being  rigidly  connected 
to  both  the  support  and  the  body,  Fig.  144. 

When  the  body  is  given  an  angular  displace- 
ment about  the  wire  as  an  axis  and  then  left 
to  itself  it  vibrates  with  a  constant  period. 
The  torque  which  produces  the  angular  displacement  obeys 
Hooke'a  law;  therefore 

G=-  IcO, 


Fig.  144. 


PERIODIC   MOTION  319 

where  k  is  a  positive  constant  which  depends  upon  the  physi- 
cal properties  of  the  wire*  The  negative  sign  indicate-  tin- 
fact  that  the  torque  and  the  angular  displacement  arc  oppo- 
sitely directed.  Substituting  this  value  of  G  in  the  torque 
equation  we  have 

ITt=-kd>  (XVII) 


or 


do* 
dt 


where  c2=  -■     But  these  are  the  typical  forms  of  the  equa- 
tion of  simple  harmonic  motion;  therefore 


2;r       n        II 


P=  —  =2 
c 


/t  (XVIII) 


V/, 


is  the  expression  for  the  period.  It  will  be  observed  that 
the  motion  is  strictly  harmonic;  consequently  there  is  no 
correction  for  finite  amplitudes. 

247.  Application  to  the  Determination  of  Moment  of  Inertia. 
—  Let  P  be  the  period  of  the  torsion  pendulum  and  P'  its 
period  after  the  body  whose  moment  of  inertia  is  desired  is 
fastened  to  the  bob  of  the  pendulum.  Further  let  /  be  the 
moment  of  inertia  of  the  bob  about  the  suspension  wire  as  an 
axis  and  V  the  moment  of  inertia  of  the  body.     Then  we  have 

p-2VI 

//  +  v 

and  P/  =  2»y-  ^ 


Therefore 


k 

P'2  -  P-  r 


and  I:  =  4 


P'--  P- 


1  Ience  if  /  is  known  both  V  and  k  may  be  determined  experi- 
mentally. 

*  Page  L78 


320  ANALYTICAL  MECHANICS 

248.  Damped  Harmonic  Motion.  —  When  a  particle  moves  in 
a  harmonic  field  of  force  which  is  filled  by  a  resisting  medium 
the  motion  of  the  particle  is  called  damped  harmonic  motion. 
The  particle  is  acted  upon  by  two  forces,  namely,  a  har- 
monic force  due  to  the  field,  and  a  resisting  force  due  to 
the  medium.  All  resisting  forces  are  functions  of  the  veloc- 
ity and  art  in  a  direction  opposed  to  that  of  the  velocity. 
But  since  in  harmonic  motion  the  velocity  does  not  attain 
great  values,  we  can  suppose  the  resisting  force  to  be  a  linear 
function  of  the  velocity.  Therefore  if  F  denotes  the  total 
force  acting  upon  the  particle  we  can  write 

F  =  —  fax  —  k->r, 

where  the  first  term  of  the  right-hand  member  represents 
the  harmonic  force  and  the  second  term  the  resisting  force. 
Substituting  this  value  of  F  in  the  force  equation  we  get 

»i --**-*»        oa%) 

A  motion  which  is  the  perfect  analogue  of  the  motion  de- 
fined by  ('([nation  (XIX)  is  obtained  when  a  rigid  body  placed 
in  a  resisting  medium  is  subjected  to  a  harmonic  torque.  The 
motion  is  defined  by  the  following  torque  equation: 

/^  =-k'd-k"u,  (XX) 

where  the  first  term  of  the  right-hand  member  represents  the 

harmonic  torque  and  the  second  term  the  resisting  torque. 

<  >n  account  of  the  perfect  analogy  between  the  two  types 

of  motion  a  discussion  of  one  of  them  is  all  that  is  necessary. 

We  will  consider  the  motion  represented  by  equation  (XX). 

/,•"  /,■' 

I    •  2a  and  —  =  b2,  then  equation  (XX)  becomes 


PERIODIC  MOTION  321 

The  last  equation  is  a  differentia]  equation  of  the  second 
order  which  can  be  solved  by  the  well-known  methods  of 
Differential  Equations.  We  will,  however,  obtain  the  solu- 
tion by  a  method  which  is  more  instructive  and  which  may 
be  called  an  experimental  method. 

It  will  be  observed  that  d  and  its  first  two  derivatives  arc 
added  in  equation  (1);  therefore  8  must  be  such  a  function 
of  t  that  when  it  is  differentiated  with  respect  to  the  time  the 
result  is  a  function  of  the  same  type.  The  only  known  ele- 
mentary functions  which  satisfy  this  condition  are  the  circu- 
lar and  exponential  functions.  But  since  circular  functions 
may  be  obtained  from  exponential  functions*  the  solution 
of  equation  (1)  may  be  expressed  in  the  form 

6  =  ae*,  (2) 

where  a  and  p  are  constants.  Replacing  6  and  its  first  two 
derivatives  in  equation  (1)  by  their  values,  which  are  ob- 
tained from  equation  (2),  we  get 

(/32  +  2  a/3  +  o2)  ae"  =  0. 

Evidently  one  or  both  of  the  factors  must  vanish.  When  ae01 
.  =  0,  0  =  0,  which  means  that  there  is  no  motion.  This  is 
called  a  trivial  solution.  When  the  other  factor  vanishes  we 
get  

j8  =  -  a  ±  vV  -  62. 

Substituting  these  values  of  /3  in  equation  (2)  we  obtain  the 

following  particular  solutions: 

8'  =  ae-{<>+**=*\ 

In  order  to  obtain  the  general  solution  we  multiply  the  par- 
ticular solutions  by  constants  and  add  them.     Hence 

0  =  ae—'  (devV-^'  +  Cfi-  y/°rr*') 

*  Sec  Appendix  Avii. 


322 


ANALYTICAL   MECHANICS 


is  the  general  solution  of  equation  (2).     Now  let  6=  0  when 
t  =  0,  then  Co  =  —  C\,     Therefore 

e=A1e—'(ev*zn:u-e-v*zr»t)}  (XXI) 

where  Ax  =  acx.     There  are  three  special  cases  which  must 
be  discussed  separately. 

Case  I.     Let  o2  =  b\  then  6  =  0  for  all  values  of  the  time. 
Therefore  this  is  a  case  of  no  motion. 

Casi  II.     Let  <r>  b2,  then  Va2-  62  is  real. 
radical  by  c  we  have 


d=Al[e-{a 

The  character  of  the  motion 
is  brought  out  by  the  graph 
of  ('(inutioii  (XXII),  Fig. 
I  1.").  The  graph  is  easily 
obtained  by  drawing  the 
dotted  curves,  which  are 
plotted  by  considering  the 
terms  of  the  right-hand 
member  of  equation  (XXII) 
separately,  and  then  adding 
them  geometrically.  It  is 
evident  from  the  curve  that 
t  he  value  of  9  starts  at  zero, 
increases    to   a   maximum, 


Denoting  this 

(XXII) 


Fig.  1 1.".. 
and    then    diminishes    to    zero 


asymptotically.     In  this  case  the  motion  is  said  to  be  aperi- 
odic «»f  dead-bi  at. 

III.     L.t  a- <b-  then  \  <r  -  (>-  is  imaginary.      Let 
s       i      i  and  V&»  -  a2  =  «.    Then  vV  -  62  =  iu.    Making 
tin-  substitution  in  equation  (XXI)  we  obtain 
e=Ale-at(eikt-c-ikt) 
=  Axe~al  -2  /sin  ul* 
=  Ac-a,smut,  (XXIII) 

*  See  Appendix  Avn. 


PERIODIC  MOTION 


:\X\ 


where  A  =  2iAi.  Equation  (XXIII)  is  the  integral  equa- 
tion of  harmonic  motion  with  the  additional  factor  e~at, 
which  is  called  the  damping  factor.  On  account  of  this  factor 
the  amplitude  of  the  motion  continually  diminishes. 

It    is  evident    from  equation  (XXIII)   that    the  motion  is 
periodic  and  has  a  period 


V62  -  a2 


(XXIV) 


The  character  of  the  motion  is  brought  out  clearly  by  the  dis- 
placement-time curve  of  Fig.  146.     A  mental  picture  of  the 


Fia.  146. 

damped  harmonic  motion  of  a  particle  may  be  formed  by  con- 
sidering the  motion  of  an  auxiliary  particle  which  moves  in 
a  logarithmic  spiral.  If  the  auxiliary  particle  describes  the 
logarithmic  spiral  of  the  figure  in  the  counter-clockwise  di- 
rect ion,  in  such  a  way  as  to  give  I  he  radius  vector  a  constant 
angular  velocity,  then  the  motion  of  the  projection  of  the 
auxiliary  particle  upon  the  "-axis  is  damped  harmonic 

The  logarithmic  spiral  may  be  used  as  an  auxiliary  curve 
in  drawing  the  graph  of  equation  (XXIII),  as  the  circle  is  used 
in  drawing  a  sine  curve. 


324  ANALYTICAL  MECHANICS 

249.  Logarithmic  Decrement. —  The  logarithm  of  the  ratio 
of  two  consecutive  amplitudes  is  constant  and  is  called  the 
logarithmic  (Urn  ment  of  the  motion.  The  amplitudes  occur 
whenever  the  relation 

tan  (orf)  =  — 

is  satisfied.  Let  the  first  amplitude  occur  at  the  instant 
t  =  U ;  then  since  the  period  of  the  tangent  is  tt,  the  times  of 
the  succeeding  amplitudes  are  given  by 

tan  (coO  =  tan  (wh  +  rnr), 


or  by  t  =  h  + 


rnr 


LO 


where  n  is  a  positive  integer.  Hence,  denoting  the  loga- 
rithmic  decrement  by  A  and  the  nth  amplitude  by  an,  we 
have 

X  =  log  — —  (by  definition) 

ocn  +  2 

=  log^e"a(<'+Hs;nMl  +  n7r) [by  (XXIII)) 

Ae~a^  +  'L^X)  sin  [a>ti  +  (n  +  2)ir] 


log 


2tt 


=  al> 

'■"i  P.  (XXV) 

Therefore  if  /  is  known  /,"  may  be  determined  from  observa- 
tion- of   /'  ;in<l  a. 

'  ( obtained  by  Betting- .   =  0. 


PERIODIC   MOTION  325 

250.    Effect  of  Damping  on  the  Period.  —  Substituting  the 
values  of  a  and  6  in  the  expression  for  the  period, 


Vy-pi 


i+^ 


=  ^0(1+^)'  (XXVI) 


where  P0  is  the  period  for  the  undamped  motion.  It  is  evi- 
dent from  equation  (XXVI)  that  the  damping  increases  the 
period. 

VIBRATIONS    ABOUT    A    POSITION    OF    EQUILIBRIUM. 

251.  Lagrange's  Method. —  In  the  various  pendulum  prob- 
lems which  we  have  discussed  the  vibrating  body  was  consid- 
ered to  be  either  a  particle  or  a  rigid  body.  These  simplifi- 
cations were  necessary  because  the  methods  we  have  used 
cannot  be  applied  conveniently  to  complicated  systems.  La- 
grange (1736-1813)  introducedinto  Dynaniii-s  a  method  which 
can  be  applied  to  any  vibrating  system.  The  following  is  a 
special  case  of  his  method  adapted  to  conservative  systems 
which  have  only  one  degree  of  freedom  of  motion. 

Express  the  potential  energy  of  the  system  as  a  function 
of  a  properly  chosen*  coordinate  q,  so  thai  when  expanded 
in  ascending  powers  of  q  the  first  power  of  9  does  qoI  appear. 
Thru  the  potential  energy  takes  the  form 

U=  A.  +  fa%  +  M  +  •  ■  •  .  xxvii) 

where  /30,  #>,  etc.,  arc  constants.    The  constant  ft  can  be 

*  It  is  shown  in  books  on  advanced  Dynamics  thai  such  a  choice  is  always 
possible. 


326  ANALYTICAL  MECHANICS 

eliminated  by  taking  the  origin  as  the  position  of  zero  poten- 
tial energy.     Thus  we  have 

U=Ptf+($zq*+ (XXVII') 

But  since  the  vibrations  are  supposed  to  be  small,  q  remains 
a  small  quantity  during  the  motion.  Therefore  the  higher 
powers  of  q  are  negligible  compared  with  q2.  Thus  neglect- 
ing the  higher  terms  we  obtain  the  following  expression  for 
the  potential  energy  of  the  system. 

£/=!/3<Z2,  (XXVIII) 

where  \  0  =  /32. 

The  kinetic  energy,  on  the  other  hand,  takes  the  form 

T=±aq2,  (XXIX) 

where  a  is  a  constant  and  q  =  --* .     But  since  the  system  is 

conservative  the  sum  of  its  dynamical  energy  remains  con- 
stant.    Therefore 

E=  T+U 

=  \  aq°-  +  \  $q\  (XXX) 

Differentiating  both  sides  of  the  last  equation  with  respect 
to  the  time, 

ag  +  /fy=0,  (XXXI) 

which  is  the  differential  equation  of  simple  harmonic  motion. 
Therefore  we  have 

q  =  a  sin  V  -  (t  +  k)  (XXXII) 

_  a 

and  P=27rV^.  (XXXIII) 

'  p 

Hence  the  main  part  of  Lagrange's  method  consists  of  select- 
ing the  coordinate  which  defines  the  position  of  the  system 
in  such  a  way  as  to  make  the  expressions  for  the  kinetic  and 
potential  energies  of  the  forms 

T=h«q\ 

U=\Pq\ 


PERIODIC  MOTION 


:;_>7 


ILLUSTRATIVE    EXAMPLES. 

1.  A  weight  which  is  suspended  by  means  of  a  helical  spring  vibrates 
in  the  gravitational  field  of  the  earth.  Find  the  expression  for  the  period, 
taking  the  mass  of  the  spring  into  account. 

Let    m   =  mass  of  the  suspended  body. 

m'  =  mass  of  the  spring. 
p  =  mass  per  unit  length  of  the  spring. 
L  =  length  of  the  spring  before  the  body  is  suspended. 
D  =  increase  in  the  length  of  the  spring  due  to  the  weight  of  the 

suspended  body. 
a  =  the  distance  through  which  the  body  is  pulled  down  in 
order  to  start  the  vibration. 

In  Fig.  147  let  0  denote  the  position  of  equilibrium,  A  the  lowest  posi- 
tion, and  B  any  position  of  the  body.  The  coordinate  in  terms  of  which 
we  want  to  express  the  energy  of  the  system  must  vanish  at  the  position 
of  equilibrium.  Therefore  we  will  define  the  position  of  the  suspended 
body  in  terms  of  its  distance  from  the  position  of  equilibrium.  The  dis- 
tance will  be  considered  as  positive  when  measured  downwards.  Let  q 
denote  this  distance   then  the  kinetic  energy  of  the  suspended 

body  equals  £  mq2.    In  order  to  express  the  kinetic  energy  of    

the  spring  in  terms  of  this  coordinate  let  as  denote  the  distance 
of  an  element  of  the  spring  from  the  point  of  suspension. 
Then  the  kinetic  energy  of  the  entire  spring  is 


1   r 


x2dm 


1  CL''  .,      , 

=  2  J,    Z?«  •"** 

=!£/> 

-If*« 

1  m' 

Hence  the  kinetic  energy  of  the  entire  system  is 

*■-!(»+*>. 

(K) 


328  ANALYTICAL  MECHANICS 

By  Hooke's  law  the  force  which  produces  the  extension  of  the  string  is 
a  harmonic  force,  that  is,  if  Q  denotes  the  force  then  Q  =  -  kq,  where  k  is 
a  constant.     Therefore  the  potential  energy  of  the  system  is 

U  =  -  CQJq 
Jo 

=  k  \  qq  dq 

Jo 
=  \kq\ 

But  Q  =  nig  when  q  =-D.     Therefore  mg  =  kD,  or  k  =  ^ .     Making 

this  substitution  in  the  expression  for  the  potential  energy  we  obtain 


u-lft. 

Therefore  the  tota 

energy  of  the  system  is 

E  =  T+  U 

1  /      ,   ?n'\  .,   .    1  mg   n 

1  tifferentiating  the 

lasl  equation  with  respect  to  t  we  obtain 

/       ,    m'\  ..    .   vw         _ 

which  is  the  equation  of  simple  harmonic  motion.     Therefore 

n  —  n   -in       '              '"■l             (f    \    f  ) 

"-"""x/(m  +  ^D(t  +  t'} 

and 

P=*'S/(1+£)T 

It  will  be  observed  that,  as  in  the  case  of  every  true  harmonic  motion, 
the  period  is  nol  affected  by  the  amplitude. 

When  the  mass  of  the  spring  is  negligible  compared  with  that  of  the 
suspended  weight  the  last  two  equations  become 

q  =  as'm\/jr(t  +  t0), 


•J- 


/§• 


Therefore  in  this  case  the  length  of  the  equivalent  simple  pendulum  equals 

■  ten  in  t  he  lengl  b  of  t  be  spring  produced  by  suspending  the  weight. 

2.   A  particle  of  mass  m  is  attached  to  the  middle  point  of  a  stretched 

j  of  natural  length  L.  modulus  of  elasticity  X,  and  of  negligible 


PERIODIC  Motion  329 

mass.     Find  the  period  with  which  the  particle  will  vibrate  when  dis- 
placed along  the  string. 

Let  U  be  the  stretched  length  of  the  string,  .1  the  area  of  its  cross-flec- 
tion, q  the  distance  of  the  particle  from  its  position  of  equilibrium,  and  'J\ 
and  T2  the  tensile  forces  of  the  two  parts  of  the  string.  Then  by  Booke's 
law  we  have 

^      x  (2"  +  V~2      .  L'-L  +  2g 
A=X 1 =  X  I  ' 

9 


^Jizitl 


S  =  X\2      V      2.xg-L-2t. 

.4  L  L 

2 

Therefore  the  resultant  force  on  the  particle  is 


where  X'  =  AX.     Hence  the  potential  energy  equals 


2X'   . 
■q: 


But  since  the  kinetic  energy  is  given  by 

1  2X' 

we  obtain  E  =  -  mq2  +  -=-  q2 

Z  Ld 

for  the  total  energy  of  the  system.     Differentiating  the  last  equation  we 
get 

mq  +  —  q  =  0, 
which  gives 

and  r*  =  ir  \         • 

3.  A  cylinder  performs  small  oscillations  inside  of  a  fixed  cylinder. 
Find  the  period  of  the  motion,  supposing  the  contact  between  the  cylin- 
ders to  be  rough  enough  to  prevenl  sliding. 


330  ANALYTICAL  MECHANICS 

Let  m  be  the  mass  of  the  vibrating  cylinder  and  a  and  b  the  radii  of  the 
vibrating  and  the  fixed  cylinders,  respectively.    Then  at  any  instant 

T  =  1  /co2, 

where  T  denotes  the  kinetic  energy 
of  the  vibrating  cylinder,  /  its  mo- 
ment of  inertia  about  the  element  of 
contact  and  o>  its  angular  velocity. 
Bui 

1  =  1  ma2, 

v 
and  co  =  -  =  (b-a)9,  FlG    14g 


where  v  is  the  linear  velocity  of  the  axis  of  the  moving  cylinder  and  6  its 
angular  velocity.    Therefore 

T  =  fm(6-a)202. 

On  the  other  hand  we  have  the  following  expressions  for  the  potential 
energy : 

U  =  mgh 

=  mg  (b  —  a)  (1  —  cos  6) 


-»-«>H'-atS~  ••)*]• 


Since  0  is  supposed  to  remain  small  all  the  time,  it  is  permissible  to  neglect 
the  higher  terms  of  9  in  the  last  expression  for  U.    Therefore  we  have 

U  =  i  mg  (b  -  a)  d\ 

Thus  both  T  and  U  are  expressed  in  forms  which  are  adapted  to  the  appli- 
cation of  Lagrange's  method. 

The  total  energy  of  the  system  is 

E  =  I  m  (b  -  a)2  02  +  |  mg  (b  -  a)  02. 
I  differentiating  the  last  equation  with  respect  to  the  time  we  obtain 

3(b-a)'d  +  2g6  =  0. 
Therefore 


*,V/i(?h)«+tt 


and 


-V 


(b  -  «) 


The  expansion  is  carried  out  by  Maclaurin'a  Theorem.   Sec  Appendix  Ai. 


PERIODIC  MoTlox  331 

When  the  contact  is  smooth  we  have 

T  =  lm  (b  -  ay-d-, 
and  U  =  |  mg  (6  -  o)  0-\ 

Therefore  0  =  a  sin  \  /  t-2—  (<  +  '«) , 

\  o  —  a 

and  P  =  2iri/^^. 

Thus  the  length  of  the  equivalent  simple  pendulum  is  (6  —  a)  when  the 
contact  is  smooth  and  — *-— — '  when  it  is  rough. 


PROBLEMS. 

1.  A  butcher's  balance  is  elongated  1  inch  when  a  weight  of  4  pounds 
is  placed  in  the  pan.  If  the  spring  of  the  balance  weighs  5  ounces,  find 
the  error  introduced  by  neglecting  the  mass  of  the  spring  in  calculating 
the  period  of  oscillation. 

2.  Find  the  expression  for  the  period  of  vibration  of  mercury  in  a 
U-tube. 

3.  If  in  the  illustrative  problem  on  p.  329  the  particle  divides  the  string 

in  the  ratio  of  1  to  n,  show  that  the  period  is  P  =  2  ir V  "'  ~    •  ^t  • 

T      //-        X 

4.  Find  the  period  of  vibration  of  a  homogeneous  hemisphere  which 
performs  small  oscillations  upon  a  horizontal  plane  which  is  rough  enough 
to  prevent  sliding. 

6.  Find  the  period  of  vibration  of  a  homogeneous  sphere  which  makes 
small  oscillations  in  a  fixed  rough  sphere. 

6.  A  particle  of  mass  m  is  attached  to  a  point  on  a  smooth  horizontal 
table  by  means  of  a  spring  of  natural  length  L.  [f  the  particle  is  pulled  so 
that  the  spring  is  stretched  to  twice  its  natural  Length  and  then  let  go,  show 

that  it  will  vibrate  with  a  period  P  =  2  (w  +  2)  y  — ,  when'  T  is  the  force 

necessary  to  stretch  the  Bpring  to  twice  its  natural  length.     The  mass  of 
the  spring  is  negligible. 

7.  Two  masses  ///,  and  tn-  are  connected  by  a  spring  o!  aegligible  mass. 
The  modulus  of  elasticity  of  the  spring  is  such  that  when  Mi  is  fixed  m2 

makes  n  vibrations  per  second.    Show  thai  when  in-  is  fixed  mi  makes 


nV^v 


brations  per  second. 


332  ANALYTICAL  MECHANICS 

8.  In  the  preceding  problem  suppose  both  of  the  particles  to  be  free 
and  show  that  they  make  n  V  " ''-±-^2  vibrations  per  second. 

9.  A  string  which  connects  two  particles  of  equal  mass  passes  through 

a  small  hole  in  a  smooth  horizontal  table.     One  of  the  particles  bangs 

vertically  while  the  other,  which  is  on  the  table  at  a  distance  D  from  the 

hole,  is  given  a  velocity  Vg~D  in  a  direction  perpendicular  to  the  string. 

Show  that  the  suspended  particle  will  be  in  equilibrium  and  that  if  it  is 

hi  i) 
slightly  disturbed  it  will  vibrate  with  a  period  of  2ir  y  —  • 

10.  The  piston  of  a  cylinder,  which  is  in  a  vertical  position,  is  in  equi- 
librium under  the  action  of  its  weight  and  the  upward  pressure  of  the  gas 
in  the  cylinder.  Show  that  when  the  cylinder  is  given  a  small  displace- 
ment it  will  vibrate  with  a  period  equal  to  2  tt  y -,  where  h  is  the  height 

of  the  piston  above  the  base  of  the  cylinder  when  the  former  is  at  its  equi- 
librium position.     Assume  Boyle's  law  to  hold. 

11.  In  illustrative  problem  2  (p.  328)  take  the  mass  of  the  string  into 
account  and  obtain  the  expression  for  the  period  of  vibration. 

12.  In  problem  6  take  the  mass  of  the  spring  into  account  and  obtain 
an  expression  for  the  period. 

13.  In  problem  7  take  the  mass  of  the  spring  into  account  and  find  the 
expression  for  the  period  of  vibrations. 

14.  In  problem  8  take  the  mass  of  the  spring  into  account  and  find  the 
expression  for  the  period. 

16.  A  pari  icle  is  placed  at  the  center  of  a  smooth  horizontal  table;  two 
particl.-  of  the  same  mass  as  the  first  one  are  suspended  by  means  of 
Btrings  of  negligible  mass,  each  of  which  passes  over  a  smooth  pulley  at 
the  middle  point  of  one  of  the  edges  of  the  table  and  is  attached  to  the 
firsl  particle.  The  particle  at  the  center  is  given  a  small  displacement 
at  righl  angles  to  the  strings.    Show  that  it  performs  small  oscillations 

with  a  period  of  2  tt  \  -  where  a  is  the  distance  between  the  two  pulleys. 

16.  A  particle  rests  at  the  center  of  a  square  table  which  is  smooth  and 
horizontal.  Four  particles  are  suspended  by  means  of  strings  each  of 
which  passes  over  an  edge  of  the  table  and  is  connected  to  the  particle  on 
the  table.  Find  the  period  with  which  the  system  will  vibrate  when  the 
le  which  is  on  the  table  is  displaced  along  one  of  the  strings.  The 
particles  have  equal  mass.    Neglect  the  mass  of  the  strings. 


PERIODIC  motion  333 

17.   A.  particle  is  in  equilibrium  at  a  point  midway  between  two  centers 

Of  attraction,  which  attract   the  particle  with  forces  proportional  to  the 
distance.    Show  that  if  the  particle  is  displaced  toward  one  of  I  he  centers 

it  will  vibrate  with  a  period  of  =  ,  where  K  and  K'  are  the  forces 

v  K  +  K' 

which  a  unit  mass  would  experience  when  placed  at  a  unit  distance  from 

each  center  of  force. 


APPENDIX  A. 


MATHEMATICAL   FORMULA. 


335 


[.   BINOMIAL  THEOREM. 

(a  +  z)»  =  a"  +  2  a-  .r  +  ?-^=^  a-**'  +  *  (*  -  1H"  "  2> 

an-3j-3  +     .    .    . 


<l+t\- 


When  x  «C  a,  and  consequently 

.r2    x3      .        „ 
— >  — ,  etc.,<Cx. 


Applying  this  theorem  to  (1  ±  x)-1  we  obtain 

—^—  =  1  -  x  +  x-  -  x3  +  •  •  • 

1  +  x 

«.'  i  _  -     pVhen  x«l,  and  consequently! 
|_x2,  a;3,  etc.,<x. 

— ^—  =  l  +  .r  +  x-  +  x3  +  •  •  • 
1  —  x 

..,  and  consequently 


[When  x  <§:  1,  ai 
.(-.  x8,  etc.,<^Cx. 


II.   QUADRATIC    FORMULA. 
If  x  .satisfies  the  quadratic  equation  ox8  +  6x  +  c  =  0,  then 


—  b  ±  V6-  -  4  (//■ 
T=  2^ 

III.  LOGARITHMIC    RELATIONS. 

(a)  lop,  ab  =  \u<r<l  +  loir/*. 

,.  .      ,       _        ,  friiis  formula  may  be  obtained  from  (a)  l>v~| 

(b)  log  an  =  n  log  a.      ,  .  .,    , 

[Jetting  b  =  a,  a-,  etc.,  until  nb  =  a  . 

(c)  log:  =  loga  — log&.    [This  follows  immediately  from  (a)  and  (b).] 

(d)  log  1  =  0.    [This  is  obtained  by  letting  6  =  </  in  (<•').] 

IV.  TRIGONOMETRIC    RELATIONS. 

sin-  X  -f-  COS'l  =  1. 
(b)       1  +  tairx  =  sfi-'.r. 

337 


338 


APPENDIX  A 


(c) 
(d) 

(e) 

(f) 
(g) 

(h) 
(i) 
(j) 


tan  (x  ±  y) 


sin  (x  ±y)  =  sin  x  cos  y  +  cos  x  sin  ( ±  y). 

cos  (x  ±  y)  =  cos  x  cos  y  —  sin  x  sin  (  ±  y) 
tanx  +  tan  (±  ?/) 
1  —  tanx  tan  (±  ?/) 

sin  2  x  =  2  sin  x  cos  x. 

cos  2x  =  cos2x  —  sin2x. 

.      n  2  tan  x 


[(f),  (g),  and  (h)  are  obtained  by  let-1 
ting  y  =  x  in  (c),  (d),  and  (e). 


sin  x  =  2  sin  -  cos  -• 
cos  x  =  cos2  -  —  sin2  -• 


2  tan 


(k)       tan  x 


Phese  may  be  obtained  by  replac- 
gxby  |  in  (f),  (g),  and  (h). 


1  -  tan-' 


(1) 
(m) 

(n) 
(o) 


sin2  x  : 

cos2  X 

.   ,x 
sin2- 


(1—  cosx). 
(1  +  cosx). 


These  may  be  obtained  easily  from  (g). 

[These  are  obtained  by  replacing  x  by" 
-  in  (1)  and  (m). 


2 

HI  -  cos  2 x). 
I  (1  +cos2x). 

1 

2 

2      2 

Angle  between  two  lines. 

(p)      cos  6  =  cos  a  cos  a'  +  cos/3cos/3'  4-  cos  7  cos  7'. 

Y     MACLAURIN'S    THEOREM. 

f(x)=J(0)+flf'(0)  +  p"(0)  +  fir(0)+-  •  • 

VI.    IMP*  >  RIANT  FUNCTIONS  EXPRESSED  AS  POWER  SERIES. 
The  following  expansions  are  carried  out  by  Maclaurin's  theorem. 

=  1  +  x.     [When  x  <£C  1,  and  consequently  x2,  x3,  etc.,  <C  x.] 


(b) 

^1!      2!       3M4!^ 

(c) 

e-«  -  1  -  '•r  -  •'"'  4-  '•''-  4-  -H 

(d) 

X          X3    .     X6         X7     . 

1!      .5!      5!      7! 

APPENDIX   A 


339 


J°         X2         X*         X6 

(e)      <*»*  =  o|-2i  +  iI-6l  + 


,  [0!  =  1] 
=  1.     [When  x<^  1,  ami  consequently  x2,  x3,  etc.,<g;x.] 


(f)       log(l+x)=y-^  + 


3!      4! 


for  -  1  <  j-  <  1 


(a) 
(b) 

(0 
(d) 


e*x  _  cos  j-  _|_  isinx. 


=  x.     [When  j<1,  and  consequently  x2, x3,  etc.,  <£. x.] 

VII.   RELATIONS   WHICH   CONNECT    EXPONENTIAL    FUNC- 
TIONS WITH   CIRCULAR  FUNCTIONS. 

"These  are  called  De  Moivre's  Theorems 
and  are  obtained  by  comparing  series  (b) 
and  (c)  of  VI  with  series  (d)  and  (e)  of 
the  same  group. 

"This  relation  is  obtained  by  subtract- 
ing (b)  from  (a). 

"This  relation  is  obtained  by  adding  (b) 
to  (a). 


cosx 


cos  x  —  i  sin  x. 


+  e- 


VIII.   HYPERBOLIC  FUNCTIONS. 

(a)  sinh  x  =  i  sin  (t'x). 

(b)  coshx  =  cos  (ix). 

These  are  the  definitions  of  the  hyperbolic  sine  and  the  hyperbolic  cosine. 
Replacing  x  by  ix  in  equations  (c)  and  (d)  of  group  VII  we  obtain  the 
following  relations  between  hyperbolic  and  exponential  functions: 

ex  -  e~x 


(c)  sinh  x 

(d)  cosh  x 


2 
e'  +  e- 


Squaring  equation  (c)  and  subtracting  it  from  the  square  of  equation 
(d)  we  obtain 

(e)       cosh2  x  —  sinh2  x  =  1. 

IX.  AVERAGE  VALUE. 

The  average  value  of  y  =  f  (x)  in  the  interval  between  x  =  Xi  and 
x  =  Xn  is  given  by 

ya——j",ydx. 

X'    —    X,<y  r, 


APPENDIX   B. 


MATHEMATICAL   TABLES. 


341 


Logarith 

ms  of  Numbers. 

,N. 

0         12, 

3    1    * 

~&~] 

6 

7 

8 

9 

P.P. 

0 

0OQ0   3010 

4771    6021 

6990 

7782 

8451 

9031 

9542 

22 

21 

1 

0000   0414   0792 

L139    L461 

1701 

20  11 

2304 

2553 

27  ss 

1  J 
4  4 

2  1 
4  2 

2 

3010:  3222 

3  12  1 

3617   3802 

3979 

4150 

4314 

1472 

4624 

6  3 

3 

4771    4914 

51  .5  1 

.',  1 85   531 5 

5441 

5563 

5682 

5798 

5911 

11 

13 

II 

2 

8  4 
in  5 
12  6 

4 

6021 

612s 

623,2 

6335   6435 

6532 

6628 

6721 

6812 

6902 

5 

0990 

7076 

716H 

7243    7324 

7lol 

7482 

7559 

70,5  1 

770!) 

IS 
17 

19 

14  7 

6 

7782 

,  85!  1 

7'.  124 

8129 

8195 

8261 

8325 

,s;;ss 

6 

s 

16.8 
18.9 

7 

8451 

8513 

8573 

8633    8692 

8751 

SSI  IS 

8865! 

8921 

8976 

SO 

19 

S 

9031 

'.tils;, 

9138 

'.Hill     U245 

9294 

03  15 

9395 

9445 

9494 

2.0 

19 

9 

It:,  12 

95!  16 

9638 

9085   9731 

0777 

osl-:; 

'.ISliS 

9912 

9956 

4   ii 
6  0 

3  8 
5  7 

10 

0000   0043 

0086 

oil's    oi7(i 

0212 

0253 

0294 

0334 

0374 

8  0 

7  6 

11 

0414    0453 

0492 

0531    0509 

(•607 

004.-, 

0082 

071!) 

0755 

9.5 
11  4 

12 

0792   0828 

0804 

0899   0934 

000!) 

1001 

103s 

1072 

1106 

13  3 

13 

L139    1173 

1200 

1239 

1271 

L303 

1335 

1507 

1399 

1430 

16  0 
is  0 
18 

15  2 
17.1 
17 

14 

1401    1492 

1523 

1553 

15s  1 

L61  1 

1644 

1075, 

1705 

1732 

15 

1761     17'.  "I 

IMS 

L847 

1 875 

10113 

1031 

1959 

1987 

2014 

1  7 

16 

2041    2068 

2095 

2122 

2148 

2175 

2201 

2227 

2250 

2270 

3.6 

3  4 

17 

2304   2330 

2355 

23S0 

2405 

2130 

2  155 

2  IsO 

250  1 

2520 

5.4 
7  2 

5  1 

6  8 

18 

255:'!    25,7 

2601 

2025 

20,-1  s 

2672 

2095 

2718 

2712 

2765 

9.0 

8.5 

19 

2788   2810 

2833 

2S56 

2878 

2900 

2!  125 

20  15 

2967 

2! is!) 

10.8 
12.6 
14  4 

10.2 
11.9 
13  6 

20 

3010   3032 

305  1 

3075 

31)00 

3118 

3139 

3100 

3181 

3201 

21 

3222   3243 

32t;:; 

32s  1 

330  1 

332! 

55  15 

3505 

3385 

3404 

16.2 
_1G 

15.3 
15 

22 

3424    3444 

3464 

34X3    3502 

3522 

55-1 1 

5,500 

3,57!  i 

55!  IS 

23 

3617    3636 

3655 

3674    3602 

3711 

3729 

3747 

3766 

57s  1 

1  ii 
3  2 

1  s 

3  0 

24 

3802  3820 

3838 

3856   3874 

3892 

3009 

5!  127 

3945 

3962 

i  5 

25 

3979   3!  i!»7 

101  1 

4031    4048 

4005 

4082 

4099 

4116 

4133 

li  4 
8  0 

6  0 

:  -, 

20 

4150    4166 

4183 

4200    4216 

4232 

4249 

120,5 

4281 

12!  »s 

9.6 

27 

4314   4330 

4346 

L362   4378 

4393 

4409 

4425 

4440 

4  150 

11  2 

12  8 

Hi  :> 
12  ii 

28 

1472    1487 

4502 

15 is    4533 

4548 

4501 

457!) 

4594 

4609 

14  4 

.     1 

2«J 

4624    4639 

165  1 

166!)    40X3 

10' is 

4715 

4728 

17  12 

4757 

14         13 

30 

1771     1786 

1800 

4814     1829 

1843 

Is.",  7 

|s71 

~  1886 

40(H) 

1   4 
2.8 
4   2 

i  a 

i 

31 

1914    4928 

4942 

1955    1969 

1983 

1997 

5011 

5( )_'  1 

5038 

32 

5051     51165 

507!' 

50!  12    5105 

5119 

5132 

51  15 

5159 

5  1  72 

5  6 

5  2 

33 

5185   5198 

51-11 

5i-  2  l    5237 

5250 

5263 

5270 

5302 

7.0 
8.4 

6  5 

7  8 

34 

5315   5328 

53  10 

5353    5300 

5378 

55!  11 

5403 

5110 

5  12s 

11  2 

9  1 
10  4 

ii  : 

35 

5!  11     5153 

5465 

517s    5490 

5502 

551  l 

5  5l' 7 

555 1 

30 

~u, 63    '>'>•'< 

5587 

5599    5611 

5623 

5617 

5658 

5i  ,7o 

12 

ll 

37 

5682   5694 

57ii5 

57 1 7 

572!  i 

57  lo 

5  ,  i  ,:  : 

5775 

5786 

1  2 

l.l 

38 

5798   5809 

5821 

5832 

5843 

5  s :,:, 

5866 

5877 

5888 

5900 

2  2 

39 

5911    5922 

5933 

5944 

5!  155 

5966 

5!  177 

5988 

5999 

6010 

i  a 

3  3 

4  4 

40 

6021    6031 

6042 

6053 

000,1 

.1075 

6085 

6096 

0107 

0117 

-  i 

5  5 

6  6 

7  7 

41 

6128   6138 

61  19 

6160    o,i7i) 

01  so 

Ol'O 

0201 

6212 

6222 

a 

6232   6243 

6253 

6263    6274 

6284 

020  1 

6304 

631  1 

6325 

9.0 

8  8 

43 

<  '<'■  ',.'<:, 

6365   6375 

6385 

6405 

6415 

6425 

9 
0  9 

'■'    ■ 
_8_ 

- 

44 

6435   6444 

6454 

6464    o(|7l 

6484 

6493 

6503 

0510 

6522 

45 

6532   6542 

6551 

6561    6571 

05  so 

05!  Ml 

6599 

6609 

6618 

1  - 

1  6 

Hi 

6628    6637 

6646 

6656   6665 

6675 

6684 

0702 

0712 

2  4 

3  2 

i 

47 

r,72l    6730 

073' I 

o,7lo    075s 

6767 

6776 

6785 

070  1 

48 

6812   6821 

6830 

6839    6848 

6857 

6866 

6875 

6884 

a 

6  4 

4  8 

49 

6902   6911 

6920 

6928    6937 

6946 

6955 

6964 

6972 

6981 

- 

■    1 

60 

6990   6998 

7007 

7(  HO    7IH-1 

7033 

70  12 

7(  )5( ) 

7059 

7067 

N. 

0          1         2 

3          4 

5 

6 

7 

8 

9 

Logarithms  of  Numbers 

N.  | 

0    1    2    3    4 

5 

6    7 

8 

9 

P.  P. 

50 

6990 

6998  7007  7016  7024 

7033 

7042  707,n 

7059 

7067 

9 

51 

7076 

7084  7093  7101  7110 

71ls 

7126  7137, 

7143 

717,2 

0.9 

52 

7160 

7168 

7177  71V,  719:; 

7202 

7210  7218 

7226 

7235 

1.8 
2.7 
3.6 

53 

7243 

7251 

7259  7267 

7277, 

728 1 

7292 

7300 

730S 

7316 

54 

7324 

7332 

7340  7348 

7356 

7361 

7372 

73SII 

7388 

7396 

4.5 

55 

7404 

7412 

7419 

7427 

7135 

7443 

7  17,1 

7  I.V.i 

7166 

7474 

5.4 
6.3 

56 

7482 

7  l'.  in 

71!  i7 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

77.7  1 

75S2 

7589 

77,!  17 

760  1 

7612 

7619 

7627 

S.l 

58 

7634 

7642 

7649  7657 

7661 

7672 

767!  I 

76sii 

7694 

7701 

8 

59 
60 
61 

7709 

7782 

7  s.-,.; 

7716 

7723  7731 

77:;^ 

77  17, 

7752 

7760 
7S32 

7767 

7774 

0.8 
16 
2.4 
3.2 

4.0 

778917796  7803 

Tmh 

7S1S 

7825 

7839 

7910 

7846 

7917 

7siiii  Tsris  7s75 

7882 

7889 

7*96 

7903 

62 

7924 

7931  7938  7945 

7952 

7959 

7966 

7973 

7980 

7987 

4.8 

63 

7993 

8000  8007  8014 

8021 

8028 

8035 

8041 

8048 

8055 

5  6 
6.4 

64 

8062 

si  169  S075  S0S2 

MIS!) 

8096 

8102 

8109 

8116 

8122 

7.2 

65 

8129 

8136  81  12  si  19 

8156 

si  62 

8169 

8176 

8182 

8189 

7 

66 

8195 

8202  8209  8215 

8222 

8228 

8235 

8241 

8248 

8254 

0.7 
1.4 
2.1 

67 

S201 

8267  8274  8280 

8287 

8293 

8299 

S306 

8312 

8319 

68 

8325 

8331  s:;:;s  8344 

8351 

8357 

8363 

8370 

8376 

8382 

2.8 

69 
70 

s:;ss 
sl7,| 

8395  8401  8407 
8457  8463  8470 

8414 

sl2l) 

S126 

8432 

8439 
8500 

8445 

3  5 
4.2 
4.9 
5.6 
6.3 

B476 

s  (S2 

8488 

S494 

8506 

71 

S5I3 

8519  8525  8531 

S537 

S5  13 

S549 

S7,7)7, 

8561 

8567 

72 

B573 

N579  N5S5  S591 

8597 

mid:; 

8(509 

8615 

8621 

S627 

6 

73 

8639  Mil:.  8651 

8657 

S663 

8669 

8675 

8681 

8686 

0.6 

74 

X692 

8698  8704  8710 

8716 

8722 

8727 

8733 

8739 

8745 

L8 

2.4 

75 

8753 

8756  8762  8768 

8774 

8779 

S7S5 

8791 

8797 

,ssi  12 

76 

8808 

SS]  1  SM'O  ,S.s25  SS31 

MS37 

ss  12 

ssls 

8854 

8859 

3.0 

77 

8865 

ss7i  ssTti  sss2  sss7 

SMI3 

8899 

8904 

8910 

8915 

3.6 
4.2 

78 

8921 

S927  S932  mi: is  mi  13 

S919 

8954 

8960 

8965 

S971 

4.8 

79 
80 
81 

8976 

V.i.sJ  s'.ls7  S993  Ml'.is 

900  1 

9009 

9017, 

9021) 

9027, 

5.4  ' 
8 

0.5 
1.0 

9031 

! i(i36  '.li'lJ  Mi H7  '.in.",:; 

905S 

9063 

9069 

9074 

7)079 

9085 

9090  9096  9101  9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

91  13  9149  9154|  9159 

9165 

9170 

9175 

9180 

9186 

1.5 
2  0 
2  5 

83 

9191 

919(5  9201  920(5 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

92  IS  9253  925S 

9263 

9269 

9274 

9279 

9284 

92S9 

3.0 
3.5 

4.0 

85 

9294 

9299  9304  9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350  9355  9360 

9365 

9370 

9375 

9380 

9385 

9390 

4.5 

87 

9395 

9400  9405  9410 

9415 

9420 

9425 

9430 

9  137, 

9440 

_4_ 

88 

9445 

9450  9455  9460 

9465 

9469 

9474 

9479 

9484 

94S9 

0.4 
0.8 
1.2 

39 

9494 

150-1  9509  9513 

97,  IS 

97,23 

!  17,2s 

'.17,33 

953S 

'  90 

9542 

9547J  9552  9557  9562 

!  17)1  il  i 

9571 

97,76 

95S1 

9586 

16 
2  0 

91 

9590  05! (5  9(500  ! II in.",  9(509 

91,1  1 

9619 

'11521 

962s 

>:. . - 

2  4 

92 

9638 

9643  9H17 

9652  9657 

9661 

9(5(5(5 

9671 

'9677, 

9680 

2.8 
3  2 
3.6 

93 

9685 

9689  9694 

9699  97n:; 

9708 

9713 

9717 

9722 

9727 

94 
96 

9731 

9731 
9782 

9711 
9786 

9745  97:,n 
[9791  979.", 

'isi'll'l 

977,!. '  9763 
9S07)  9X09 

976S 
9814 

9773 

9818 

9823 

9827 

9836  9841 

984S 

9850  9854 

9859 

9S63 

a-  =  3.1416 

97 

9868 

9872 

9877 

9881  9886 

9MIO 

9894  9899 

9903 

!  191  IS 

log7r  =  o  4971 
<?  =  2  7is:i 

98 

9912 

9917  9921 

'.r.rji;  9930 

9934 

9939  9943 

9948 

9952 

log  e  =  0.4343 

99 

9956  9961  9965 

997s 

9983  9987 

9991 

9996 

ii  =  loir,  X 

100 

ihkhi  0004  oT)09  0013  0017 

1  jj !'_"_ 

0026!  0030 

0035 

0039 

=  2 . 2112  Iorio  * 

N. 

0    1    2    3    4 

5 

6    7 

8 

9 

Trigonometric  Functions. 
0°-22  .5. 


Degree 

— 

Sine, 

Tangent. 

( iotangent 

Cosine. 

.... 

It  IMMtO 

0   U(MM) 

oo 

1  onoo 

90.0 

ii  5 

0.0087 

0.0087 

1  1  1  5887 

1  0000 

so   :, 

10 

0  (117.-) 

It  (117.-. 

57.29(H) 

0  9998 

so   1) 

15 

ii  0262 

ii  0262 

38  1885 

it  ooo7 

88.5     , 

2.0 

ii  0349 

n  0349 

us  6363 

,,  .,.,.,1 

SN     II 

2.5 

it  0436 

0.0437 

22  !MttS 

o  9990 

s7  :, 

3.0 

'    0.0523 

it  0524 

19.0811 

0.9986 

s7  ii 

3.5 

0.0610 

ii  0612 

Hi  3499 

it  9981 

so  .-, 

4.0 

(1   ItOOS 

0.0699 

1  1  3007 

It   '.1070 

86.0 

4.5 

()  07V. 

(t  ()7S7 

12.7062 

0   0000 

85.5 

5.0 

0.0872 

()  0875 

li    i:;oi 

0.9962 

85.0     ■ 

5.5 

0.0958 

0.0963 

10.3854 

ii  9954 

1     84.5 

6.0 

0.1045 

0.1051 

9  5144 

It  00  17, 

;     84.0 

6.5 

0.1132 

0  1139 

S  7700 

II   O'.IOO 

83.5     i 

7.0 

0.1219 

0.1228 

8  I  1 13 

0  0027, 

83.0 

7.5 

0.1305 

0.1317 

7.5958 

0.9914 

82.5 

8.0 

0.1392 

0.1405 

7.1154 

0.9903 

82.0 

8.5 

0.1478 

1)    1  lit:. 

6.6912 

0   OVID 

81.5 

9.0 

it   L564 

0.1584 

6.3138 

o  OS77 

81.0 

9.5 

0.1650 

0.1673 

5  9758 

0.9863 

80.5 

10.0 

0.1736 

0.1763 

5.6713 

II    '.ISIS 

80.0 

10.5 

0.1822 

0.1853 

5.3955 

0.9S33 

79.5 

11.0 

0.1908 

0.1944 

5   1446 

n  9816 

79.0 

j     115 

0.1994 

0.2035 

I  9152 

0.9799 

7  s  :, 

12.0 

0.207!) 

0.2126 

4  70  10 

0.9781 

7s  It 

12.5 

0.2164 

0.2217 

1  .-,107 

0.9763 

77  5 

13.0 

0.2250 

0.2309 

4.3315 

0  0711 

77  ii 

13.5 

0.2334 

It   2101 

4.1653 

ii  0721 

76  5 

14.0 

()  2419 

ii  2493 

1  nlits 

ii  9703 

70  H 

14.5 

0.2504 

o  2586 

:;  soo7 

ii  9681 

77,  5 

15.0 

it  2.-.S.S 

(i  2679 

3.7321 

o  9659 

77,  0 

IS  5 

0.2672 

0.2773 

:;  6059 

o  9636 

71  5 

16.0 

0.2756 

ii  2867 

3   874 

ii  9613 

71  it 

16.5 

ii  2840 

ii  2962 

:;  3759 

ii  9588 

, . ;  5 

17.0 

it  2924 

it  3057 

3  27no 

ii  9563 

7:;  i» 

17  :, 

0.3007 

ii  3153 

:;  i7io 

0  9537 

72  5 

is  0 

ii  3090 

it  3249 

:;  H777 

It  07,11 

72  n 

In  5 

ii  3173 

ii  3346 

2  9887 

n  'tis:; 

71   5 

LQ  it 

it  3256 

o  3443 

2  on  12 

it  9455 

71   it 

19  5 

ii  3338 

ii  3541 

2  8239 

H  0  120 

7it  5 

20  it 

it  3420 

ii  3640 

2  7  17:. 

7"  ii 

•Jit  5 

it  : :.-,(»_' 

0  3739 

2  0710 

21  0 

it  3584 

ii  3839 

2  6051 

69  ii 

21  :. 

(t  3665 

ii  9304 

22  it 

ii  3746 

ii   K)40 

2    177,1 

ii  ''272 

OS  II 

n  3827 

it   11  12 

2    1112 

, 

Degree. 

67  .5 

90  . 

Trigonometric  Functions. 
22°.5-45°. 


Degree. 

Sine. 

T;in-ent. 

Co  tangent. 

Cosine. 

22.5 

0.3827 

0.4142 

2.4142 

0.9239 

67.5 

23.0 

0.3907 

0.4245 

2.3559 

0.9205 

67.0 

23.5 

u  39S7 

II    13  IS 

2.2998 

0.9171 

66.5 

24  0 

0   H  m  17 

ii   1452 

2  2460 

0.9135 

66.0 

24   5 

n    11  17 

0   1557 

2   L943 

0.9100 

65.5 

25.0 

0.4226 

0.4663 

2.1445 

0.9063 

65.0 

25  5 

n   B05 

0.4770 

2.0965 

0.9026 

64.5 

!     26  0 

ii   i:m 

0.4877 

2.0503 

II  suss 

64.0 

26  5 

0.4462 

ii   1986 

2.0057 

0.8949 

63.5 

27.0 

0.4540 

0.5095 

1.9626 

(1  Mill) 

63.0 

27.5 

n   1617 

0.5206 

1.9210 

0.8870 

62.5 

28.0 

ii   1695 

0.5317 

1.8807 

0.8829 

62.0 

!     28.5 

0   1772 

0.5430 

1 .  s  1 1  s 

n  S7ss 

61.5 

29.0 

0    IMS 

0.5543 

1    Slll'l 

0.S746 

61.0 

29.5 

0  4924 

0.5658 

1.7675 

0.8704 

60.5 

j     30.0 

0.5000 

(i  5774 

1.7321 

0.8660 

60.0 

30  5 

0  5075 

0.5890 

1.6977 

0.8616 

59.5 

31.0 

0  5150 

0.6009 

1.6643 

0.8572 

59.0 

31.5 

0.5225 

0.6128 

1.6319 

0.8526 

58.5 

32.0 

0.5299 

0.6249 

1.6003 

(1  X4S0 

58.0 

,     32.5 

0.5373 

0.6371 

1.5697 

0.8434 

57.5 

33  0 

0  r>\w> 

0.6494 

1.5399 

0.8377 

57.0 

33.5 

0.5519 

0.6619 

1.5108 

0.8339 

56.5 

:     34.0 

0.5592 

0.6745 

1.4826 

0.8290 

56.0 

34  5 

0.5664 

0.6873 

1.4550 

0.8241 

55.5 

35.0 

0.5739 

0.7002 

1.4281 

0.8192 

55.0 

1     35.5 

0.5807 

0.7133 

1.4019 

n  81  ii 

54.5 

36.0 

i)  5878 

0.7265 

1.3764 

II    Ml! Ill 

54.0 

36.5 

n  .V.ils 

0.7400 

1.3514 

0.8039 

53.5 

37.0 

0.6018 

0.7536 

1.3270 

0    7!IMi 

53.0 

37.5 

II  (iOSS 

0.7673 

1.3032 

0.7934 

52.5 

38  ii 

0.6157 

0.7813 

1.2799 

0.7880 

52.0 

38  5 

0.6225 

0.7954 

1.2572 

0.7826 

51.5 

39.0 

ii  6293 

0   Ml! IN 

1.2349 

0.7771 

51.0 

39.5 

0  6361 

0.8243 

1.2131 

0.7716 

50.5 

40.0 

ii  6428 

0.8391 

1.1918 

0.7660 

50.0 

40.5 

0  6494 

ii  8541 

1   1708 

0.7604 

19.5 

11  ii 

n  6561 

ii  8693 

L.1504 

n  7;.  17 

49.0 

n  5 

ii  6626 

n  8847 

1 . 1303 

0.7490 

48.5 

12  ii 

n  6691 

0  9004 

1 . 1 106 

0.7431 

4S.0 

12  5 

n  6756 

0.9163 

1.0913 

0.7373 

47.5 

i:;  0 

(i  6820 

ii  9325 

1.0724 

0.7314 

47.0 

i:;  :, 

ii  6884 

ii  9490 

1  0538 

0.7254 

46.5 

11  II 

n  6947 

ii  9657 

1.0355 

0.7193 

46.0 

11  .-. 

ii  7009 

n  9827 

1.0176 

0.7133 

45.5 

r,  0 

ii  7U7I 

1  0000 

1.0000 

0.7071 

45.0 

Tangent. 

Bine. 

Degree. 

45°-67°.5. 


Exponential  Functions. 


X 

e* 

X 

tr 

X 

e~" 

0.00 
.01 
.02 
.03 
.04 
.05 
.06 
.07 
.08 
.09 
.10 
.12 
.14 
.16 
.18 
.20 
.22 
.24 
.26 
.28 
.30 
.32 
.34 
.36 
.38 
.40 
.42 
.44 
.46 
.48 
.50 
.52 
.54 
.56 
.58 
.60 

1.000 

1    (HHI 

0.60 
.62 
.64 
.66 
.68 
.70 
.72 
.74 
.76 
.78 
.80 
.82 
.84 
.86 
.88 
.90 
.92 
.94 
.96 
.98 

1.00 
.1 
.2 
.3 
.4 

1.5 
.6 
.7 
.8 
.9 

2.0 
.1 
.2 
.3 
.4 

2.5 

1.822 

0.549 

2.5 
6 

7 

.8 
.9 

3.0 

.2 

.3 
.4 

3  5 

ii 
.7 
.8 
.9 

12    Is 

0  0821 

1.010 
1.020 
1.030 
1.041 

(1   -.CHI 

0.980 
0.970 

II    '.Hil 

1  859 

1    V.Hi 

1.935 
1.974 

0.538 

(1  527 
0.517 

0  ;,ii7 

13   16 

11   ss 

ir,  ii 
18.17 

ii  i>7  13 

0  0672 
ii  0608 
o  0550 

II    II  IMS 

1.051 

0.951 

2.014 

0.497 

20  0! » 

1.062     0.942 
1.073     0  932 
l  itv;     (i  923 
1.094     0.914 

2  054 

•_-  n«  in 
2  138 
2.181 

0.487 
0.477 

(i   His 
0.458 

22.20 
21   53 
27    11 
29.96 
33.12 

oloooo 

tO    4*  CD  OCO 

1.105     0.905 

2.226 

0.449 

1.127 
1.150 
1.174 
1.197 

0.N.S7 

0.869 
0.852 
0.835 

2  -Tn 
2.316 
2.336 

2    III 

0    (111 
0.432 
0.423 
0.415 

36.60 
40.45 

II  70 

ID    in 

i»  0273 

0  0217 
0  0224 
ii  0202 

1.221 

0.817 

2.460 

0.407 

4.0  ]  54.60  ]  0.0183 

1.246 
1.271 
1.297 
1.323 

0.803 
0.787 
0.771 
0.756 

2.507 
2.560 
2.612 
2  mil 

0.399 
0.391 

ii  :;s:; 
II  375 

.1 

.2 

.3 

4.5 

tin  31      0  0166 
66  69     0.0150 
73.70     0.0136 
90.02  i  o  oill 

1.350 

0.741 

2  71S 

0.368 

5.0    MS    1      n  00674 

1.377 
1.405 
1.433 
1.462 

0.726 
0.712 
0.698 
0.684 

3.004 
3.320 
3.669 
4.055 

0.333 
0.301 
0.273 

11.217 

6.0 

7.0 
8.0 
9.0 
10 

w    1 

7T/3 

x/2 
3*-/4 

7T 

3x/2 
2* 

5jt/2 
3tt 

103  4 
1097. 
2981. 
8103. 
22026. 

0  00248 

0.000912 
0.000335 
o  00012:; 
0000454 
o  4560 
0.3513 
0.2079 
0.0948 

1.4!  12 

0.670 

4.482 

ii  223 

1.522 
1.553 
1.584 
1.616 

u.  <■,-,: 

0:644 
0.631 
0.619 

4.953 

5    17  1 
6.050 

(i  lis.i 

0.202 
0.183 
0.165 

ii    150 

2.193 
2.847 
4.810 
10.55 

1.649 

0.607 

7.389 

(i    135 

23    1  I 

0.0432 

1   Os2 
1.716 
1.751 

1  7sf. 

(I  .7.1.-, 
0.583 
0.571 

ii  :,.;h 

,N    H ill 

9.025 
'.i  974 
11.02 

0.122 
0.111 
0.100 
0.907 

111   3 

535  5 

2576 

123!  12 

11  OOSitS 
0  oii|s7 
ii  000388 

1.822 

(»  51!) 

12.18 

0.0821 

286751 

log, 


2.292  log 


INDEX. 

The  numbers  refer  to  pages. 


Acceleration,  89. 
angular,  97. 
normal,  92. 

radial,  95. 
tangential,  92. 

transverse,  95. 
Action  and  react  ion,.  15. 

angular,  35,  218. 

law  of,  15,  39,  101,  197,  218. 

linear,  35,  101. 

types  of,  35. 
Adiabatic  compression,  176. 

elasticity,  177. 
Amplitude,  13G,  29S. 

correction  for,  311. 
Atwood's  machine,  119. 
Average  value,  142,  301. 
Axis,  instantaneous,  32,  228. 

of  rotation,  30. 

of  spontaneous  rotation,  279. 

Belts,  67,  71. 
Binomial  theorem,  337. 
Boyle's  law,  17ii. 

Cable,  dip  of,  62. 
length  of,  66. 

( 'at diary.  »'••">. 

( Sentrodes,  32. 

(  !ircle,  auxiliary,  209. 

( 'ollisiou.  244. 

Comparison   of  translation   and   rota- 
tion. 222. 

Composition  of  harmonic  mot  |< 
equal  periods,  303,  i"  '<■ 

Configuration,  193. 

standard,   194. 

Contact ,  elastic,  -'if.. 
inelastic,  247. 


( !oordinates,  spherical,  1  17. 
(  Hid-,  equilibrium  of,  60. 
Couple,  35. 

arm  of,  37. 

plane  of,  37. 

Damping  factor,  323. 
Degrees  of  freedom,  14,  39. 
De  Moivrc's  theorem,  339. 
Dimensions,  75. 

of,  see  Units. 
Dip  of  cable,  62. 
Displacement,  77. 

angular,  86. 

in  S.  H.  M.,  298. 

most  general,  34. 

screw-,  34. 

virtual,  181. 
Dynamical  energy,  96. 

conservation  of,  96. 
Dynamics,  2. 
Dyne,  108. 

Efficiency  of  a  blow,  251. 
Elastic  limn.  17:;. 
.  246 

Elasticity,  adiabatic,   177. 

isothermal,  177. 

modulus  of,  17:;. 

perfi  ei.  246. 

Bhearing,  178. 
Energy,  186. 

conservation  of.  196. 

degradation  of.  203. 

equation,  197. 

kinetic,  lsii,  lss. 

lost  in  collision,  _'  17.  2 19. 

method,   2! 


349 


350 


INDEX 


Energy,  potential,  186,  194. 

transformation  of,  186. 
Epoch,  300. 

angle,  300. 
Equilibrium,  ;t  special  case  of  motion, 
L73. 

and  potential  energy,  208. 

conditions  of,  16,  W,  208. 

of  a  pari  ick,  16. 

of  ri^id  bodies,  40. 

of  flexible  cords,  chains,  belts,  etc., 

stability  of,  208. 

stable,  unstable  and  neutral,  210. 
Erg,  167. 

Expansion  into  series,  338. 
Exponential,  curve,  66,  70. 

fund  inn.-.  339,  -117. 

Field  intensity,  212. 
Fields  of  force,  see  Force. 
Flexibility,  60. 
I  orce,  1 1.  i">. 

and  momentum,  239. 

central,  L26,  283. 

central  fields  of,  283. 

conservative,  193. 
coplanar,  11,  48. 
dissipative,  194. 
equation,  106. 
external,  15. 
fields  of,  203,  210,283. 
{notional,  21. 
in  a  field,  204. 

internal,    l.">,    12. 

moment  of,  38. 

inian,  210. 
aonconservative,  19 1. 
iltant,  ion. 

resultant  of,  25,   17,   19. 

transmissibility  <>f,  12. 

virtual,   181. 

.i.e.  mathematical, 
■  on,  angle  of,  21. 


Friction,  belts,  67. 
coefficient  of  rolling,  56. 
coefficient  of  sliding,  22. 
couple,  57. 
kinetic,  22. 

laws  of,  22. 

on  journal  hearings,  51. 

on  pivot  hearings,  53. 

rolling,  56. 

sliding,  21. 

static,  22. 
Fund  ions,  circular,  339. 

exponential,  339. 

hyperbolic,  339. 

trigonometric,  337. 
Fundament al  magnitudes, 


71. 


Gravitational,  acceleration,  109,  314. 

constant,  210. 

law,  210,  289. 
Guide  plane,  31. 
Gyration,  radius  of,  155. 

Harmonic,  see  Motion. 
Homogeneous  equation,  76. 
Hooke's  law,  173. 
Hyperbolic  functions,  339. 

Impact,  249. 

oblique,  258. 
Impulse,  angular,  265. 

linear,  238. 

of  compression,  245. 

of  restitution,  245. 
Impulsive  reaction,  279. 
Inertia,  100. 

moment  of,  152. 
Introduction,  1. 
Isolated  system,  195. 
hot  hernial  compression,  175. 

Joule,  167. 

Kepler's  laws,  294. 
Kilowatt,  L91. 


i  m  )  i :  x 


351 


Kinematics,  2. 

Kinetic  reaction,  angular,  218. 

lunar,  100. 

measure  of  angular,  220 

measure  of  linear,  103. 

normal,  106. 

tangential,  105. 
Kinetics,  2. 

Lagrange's  method,  325. 
Lever  arm,  39. 
Logarithmic  decrement,  324. 
Logarithmic  tables,  343. 

Maclaurin's  theorem,  338. 
Mass,  71.  L02. 

and  weight,  109. 

center  of,  140. 

comparison  of,  110. 

of  a  planet,  293. 
Mechanical  advantage,  183. 
Mechanics,  scope  of,  1. 

divisions  of,  2. 
Moment  of  force,  3S. 
Moment  of  inertia,  L52. 

experimental  determination  of,  319. 

experimental  definil  ion  of,  220,  221. 

mathematical  definition  of,  L52. 

theorems  on,   1")  I. 
Momentum,  angular,  266. 

conservation  of  angular,  268. 
conservation  of,  241. 
force  and,  239. 

linear,  239. 

of  a  system,  241. 
moment  of,  266. 

torque  and  angular,  267. 

Motion,  al)out  a  fixed  axis,  224. 
aboul  instantaneous  axes,  228. 
along  an  inclined  plane,  16. 
analysis  of,  73. 
aperiodic,  322. 
damped  harmonic,  320. 
dead  beat,  322. 
elliptic  harmonic,  306. 
equations  of,  1 1 1. 


Motion,  in  resisting  media,  130. 
of  center  of  mass,  2  I-' 
of  falling  bodies,  LI  I,  L27. 

of  two  gravitating  particles,  2*7. 

of   projectiles,     120. 

of  a  particle,  100,  113. 
of  a  rigid  body,  30,  218. 

of  rotation,  30. 
of  translation,  30. 
periodic,  L33,  297. 
relativity  of,  7:;. 
relative  to  center  of  mas.-.  272. 
simple  harmonic,  133,  297. 
uniformly  accelerated,  113. 
uniform  circular,  125. 
uniplanar,  31,  228. 
where  mass  varies,  !'.">:;. 

Newtonian,  field  of  force,  210. 

law  of  force,  210. 

potential,  211. 
Notation,  3. 

table  of,  xi. 

Orbits,  equation  of,  285. 

types  of,  290. 
( Oscillations,  small,  325. 

Parallelogram  method,  5. 
Particle,   1  1. 

auxiliary,  2'.".'. 
Pendulum,  ballistic,  271. 

bifilar,  315. 

compound,  308. 

equivalent  simple,  312. 

physical,  308. 

reversible,  21 1. 

simple,  oil. 

torsional,  318. 
Percussion,  center  of.  279 
Period,  136,  292,  298 

effect  of  damping  on  thi 

half  Value,  66,   70. 

Periodic  function.  298. 

I 


352 


INDEX 


Potential,  due  to  a  particle,  211. 

due  to  any  distribution,  212. 

field,  212. 

Newtonian,  210. 
Power,  191. 
Pressure,  172. 

volume-diagram,  175. 


Torque,  representation  of,  37. 

resultant,  219. 
Torsion,  172. 
Triangle  method,  5. 
Trigonometric  functions,  345. 

relations,  337. 
Tycho  Brahe,  294. 


Quadratic  formula,  337. 

Radius  of  gyration,  155. 
Range,  122. 

for  sloping  ground,  123. 
Reaction,  see  Action. 

frietional,  21. 

normal,  21. 

of  avis,  276. 

total,  21. 
Reference  system,  73. 
Resiliency,  246. 
Restitution,  coefficient  of ,  246. 

impulse  of,  245. 
Rigid  body,  30. 
Rigidity,  torsional,  177. 
Rout h's  rule,  163. 

Dip. 
Bcalars,  •'{. 
Screw  motion,  34. 

Shear.    172. 

Simplification  of  problems,  60. 
Speed,  77. 

Stability,  criterion  of,  208. 
Statics,  '-'. 

Strain,   17_'. 
17.'. 

Suspension  bridge,  61. 

Tension,  17.'. 
Torqui 

and  angular  momentum,  267. 

equation,  221, 

in  a  Held,  206. 

Ill' 


Unit,  angle,  87. 

angular  acceleration,  97. 

angular  velocity,  87. 

derived  magnitudes,  74. 

energy,  167. 

force,  108. 

fundamental  magnitudes,  74. 

impulse,  angular,  266. 

impulse,  linear,  239. 

length,  74. 

mass,  75. 

momentum,  angular,  266. 

momentum,  linear,  239. 

of  acceleration,  91. 

power,  191. 

systems  of,  76. 

time,  74. 

torque,  37,  221. 

velocity,  78. 

weight,  76,  109. 

work,  167. 

Vectors,  3. 

addition  of,  4. 
analytical  method,  10. 
difference  of  two,  7. 
graphical  method,  9. 
multiplication  by  scalars,  11. 
origin  of,  3. 

parallelogram  method,  5. 
representation  of,  3. 
resolution  of,  8. 
resultant  of,  5,  9. 

Bubl  faction  of,  7. 
terminus  of,  3. 

Velocity,  77. 
angular,  87. 


INDEX 


353 


Velocity,  componenta  of,  79. 

from  infinity,  129,  291. 

limiting.    131,  233. 

orbital,  290. 

radial,  81. 

relative,  83. 

transverse,  81  , 
Vibrations,  325. 
Virtual,  displacement,  181. 

force,  181. 

work,  180. 

Watt,  191. 
Weight,  109. 
and  mass,  109. 


Work    done,    againsl     conservative 
forces,  194. 
againsl  frictional  forces,  185. 
againsl  gravitational  forces,  166. 
against  kinetic  reaction,   185,  186. 
by  ;i  force,  164. 
by  a  torque,  169. 
by  components  of  force,  167. 
in  compressing  Quids,  170. 
in  stretching  a  string,  174. 
in  twisting  a  rod,  179. 
results  of,  185. 
virtual,  180. 

Young's  modulus,  173. 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall. 


f  War,64rt/W 


i\Lo  O    f    [^ 

fEfri-6-WclpM 


;APR301367    9 — 


STACKS 


"MAR— 5-13H5- 


, RFC  CIRC    APR?  7  198! 


peceivtw 


V67-4Vt* 


I   DEI 


14 — _ 


LD21A-40m-n.'ii:J 
(E1602slO)476B 


General  Library 
University  of  California 

Berkeley 


$s 


GENERAL  LIBRARY    U.C.  BERKELEY 


'S~S 


JS 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


